Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
trying to solve $\sqrt{\cos(x)-2\cos(2x)}+\sqrt{2}\cos(2x)=0$ The equation is
$$\sqrt{\cos(x)-2\cos(2x)}+\sqrt{2}\cos(2x)=0$$
The system is
$$
\begin{cases}
\cos(x)-2\cos(2x)=2\cos^2(2x) \\
-\sqrt{2}\cos(2x)\ge 0 \iff \cos(2x)\le 0
\end{cases}
$$
The equation:
$$\cos(x)-2(2\cos^2(x)-1)=2(2\cos^2(x)-1)^2$$
$$\cos(x)-4\cos^2(x)+2=2(4\cos^4(x)-4\cos^2(x)+1)$$
$$\cos(x)-4\cos^2(x)+2=8\cos^4(x)-8\cos^2(x)+2$$
$$8\cos^4(x)-4\cos^2(x)-\cos(x)=0$$
But what to do next?
If I factor out $\cos(x)$, I get a non-factorizable third-degree polynomial in parentheses
$$\cos(x)(8\cos^3(x)-4\cos(x)-1)=0$$
The problem with its solution (upside-down) from the textbook
| HINT:
$$8\cos^3x-4\cos x-1=4\cos x(2\cos^2x-1)$$
$$\implies2(\cos3x+\cos x)=1\iff4\cos x\cos2x=1$$
If $\sin x=0,4\cos x\cos2x\ne1\implies\sin x\ne0$
$$4\cos x\cos2x=1\iff\sin x=2(2\sin x\cos x)\cos2x=2\sin2x\cos2x=\sin4x$$
$$\sin4x=\sin x\implies4x=n\pi+(-1)^nx$$ where $n$ is any integer
Also, $\cos2x\le0\implies2m\pi+\dfrac\pi2\le2x\le2m\pi+\dfrac{3\pi}2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1413812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Simultaneous orthogonal diagonalization of two matrices Let $A=\begin{pmatrix}
1 & -2\\
-2 & 5
\end{pmatrix}$ and $B=\begin{pmatrix}
-3 & 6\\
6 & -10
\end{pmatrix}$. Obviously $A$ is positive-definite and thus we can simultaneously diagonalize $A$ and $B$. It's easy to check that $P=\begin{pmatrix}
1 & 2\\
0 & 1
\end{pmatrix}$ is invertible and diagonalizes $A$ and $B$ such that $P^t A P=I$ and $P^t B P=\text{diag}(-3,2)$. But is there an orthogonal matrix $Q$ such that $Q^t A Q$ and $Q^t B Q$ are diagonal? How can I possibly determine whether such a matrix exists?
| Simultaneous congruence via an orthogonal matrix implies in particular simultaneous orthogonal diagonalization and in particular, simultaneous diagonalization. Assume that there exists an orthogonal matrix $Q$ such that $Q^tAQ = \mathrm{diag}(\lambda_1, \lambda_2)$ and $Q^tBQ = \mathrm{diag}(\mu_1, \mu_2)$. Since $Q^t = Q^{-1}$, we must have that $\lambda_1,\lambda_2$ are the eigenvalues of $A$ and the columns of $Q$ are an orthonormal basis of eigenvectors (with respect to the standard inner product on $\mathbb{R}^2$) of $A$ corresponding to $\lambda_i$. Similarly, $\mu_1,\mu_2$ are the eigenvalues of $B$ and the columns of $Q$ must form an orthonormal basis of eigenvectors of $B$ corresponding to $\mu_i$. The matrices $A$ and $B$ will be simultaneously congruent to diagonal matrices via an orthogonal matrix if and only if you can find an orthonormal basis of $\mathbb{R}^2$ that consists of eigenvectors both of $A$ and $B$.
The characteristic polynomial of $A$ is $\chi_A(t) = t^2 - 6t + 9$ and so $\{\lambda_1, \lambda_2 \} = \{3 - 2\sqrt{2}, 3 + 2\sqrt{2}\}$. The corresponding eigenspaces are
$$ V_{3 - 2\sqrt{2}} = \mathrm{span}\{(1 + \sqrt{2}, 1)^t\}, V_{3 + 2\sqrt{2}} = \mathrm{span}\{(1 - \sqrt{2}, 1)^t\}$$.
Since
$$ \begin{pmatrix}
-3 & 6\\
6 & -10
\end{pmatrix} \begin{pmatrix} 1 + \sqrt{2} \\ 1 \end{pmatrix} = \begin{pmatrix} 3 + 6\sqrt{2} \\ -4 + 6\sqrt{2} \end{pmatrix} \neq (-4 + 6\sqrt{2}) \begin{pmatrix} 1 + \sqrt{2} \\ 1 \end{pmatrix}$$
we see that an eigenvector of $A$ won't be an eigenvector of $B$ and thus simultaneous congurence via an orthogonal matrix is impossible.
Alternatively, you can prove that two symmetric real matrices are orthogonally diagonalizable simultaneously if and only if they commute. Since $AB \neq BA$, this shows that $A$ and $B$ can't be diagonalized simultaneously.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1414002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Finding the definite integral of a trigonometric expression Find the integral of $$ \int_0^{\frac{\pi}{2}}{{\sqrt{\sin(2\theta)}} \cdot \sin(\theta)d\theta}$$
I got $$I=\int_0^\frac{\pi}{4}{\sqrt{\sin(2\theta)} \cdot (\sin(\theta)+\cos(\theta))d\theta}$$
But, I'm stuck here.
| The integral
$$I = \int_{0}^{\pi/2} \sqrt{\sin(2\theta)} \, \sin\theta \, d\theta$$
is evaluated by making use of the Beta function. This is seen as follows.
\begin{align}
I &= \int_{0}^{\pi/2} \sqrt{\sin(2\theta)} \, \sin\theta \, d\theta \\
&= \sqrt{2} \, \int_{0}^{\pi/2} \sin^{3/2}(\theta) \, \cos^{1/2}(\theta) \, d\theta \\
&= \sqrt{2} \cdot \frac{1}{2} \, B\left(\frac{3}{4}, \frac{1}{4}\right) \\
&= \frac{\Gamma\left(\frac{1}{4}\right) \, \Gamma\left(\frac{3}{4}\right)}{4 \, \sqrt{2}} = \frac{\pi}{4}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1416231",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
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Simple Trig Equations - Why is it Wrong to Cancel Trig Terms? In the following problem, I first did it using a cancellation of $sin^2\theta$, working shown below, which gave the wrong answer. Having looked at the question again, I saw it could be solved by factoring, working again below.
My question then, is why is it wrong to cancel the $\sin^2\theta$ term - the algebra seems correct to me?
The Problem
Solve for $\theta$ in the interval $0 \le \theta \le 360$
$$4\sin\theta = \tan\theta$$
My Solution
$$4\sin\theta = \frac{\sin\theta}{\cos\theta}$$
$$4\sin\theta \cos\theta = \sin\theta$$
Squaring, and substituting, using the identity $\cos^2\theta = 1 - \sin^2\theta$
$$16\sin^2\theta(1-\sin^2\theta) = \sin^2\theta$$
$$1-\sin^2\theta = \frac{\sin^2\theta}{16\sin^2\theta}$$
$$1-\sin^2\theta = \frac{1}{16}$$
Rest of working to final answer omitted.
Correct Solution
$$4\sin\theta = \frac{\sin\theta}{\cos\theta}$$
$$4\sin\theta \cos\theta - \sin\theta = 0$$
$$\sin\theta(4\cos\theta - 1) = 0$$
Rest of working to final answer omitted.
In sum, why is it wrong to cancel as I did first time around - why must these problems be solved by factoring as in the correct solution?
| Here's a simpler way to get your answer. Remember what $sin(\theta)$ and $tan(\theta)$ actually represent on a right triangle. For a given right triangle, let $a$ be the length of the adjacent side, $b$ be the length of the opposite side, and $c$ be the length of the hypotenuse. This means:
$sin(\theta) = \frac{a}{c}$ and $tan(\theta) = \frac{a}{b}$
Substituting, your equation becomes:
$4\frac{a}{c} = \frac{a}{b}$
Cancelling the $a$ and rearranging gives:
$\frac{b}{c} = \frac{1}{4}$
Using the definition of cosine gives:
$cos(\theta) = \frac{1}{4}$
And
$acos(\frac{1}{4}) = \theta$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1416619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 4
} |
Matrix exponential: $\begin{pmatrix} 0 & 1 \\ -4 & 0 \end{pmatrix}$ It is asked to calculate $e^A$, where
$$A=\begin{pmatrix} 0 & 1 \\ -4 & 0 \end{pmatrix}$$
I begin evaluating some powers of A:
$A^0= \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\; ; A=\begin{pmatrix} 0 & 1 \\ -4 & 0 \end{pmatrix} \; ; A^2 = \begin{pmatrix} -4 & 0 \\ 0 & -4\end{pmatrix} \; ; A^3 = \begin{pmatrix} 0 & -4 \\ 16 & 0\end{pmatrix}\; ; $
$ A^4=\begin{pmatrix} 16 & 0 \\ 0 & 16\end{pmatrix},\; \ldots$
I've noted that, since
$$e^A = \sum_{k=0}^\infty \frac{A^k}{k!}$$
we will have the cosine series at the principal diagonal for $\cos(2)$. But couldnt get what we will have in $(e^A)_{12}$ and $(e^A)_{21}$.
Also, we know that if $B=\begin{pmatrix} 0 & \alpha \\ -\alpha & 0 \end{pmatrix}$, then $e^B = \begin{pmatrix} \cos(\alpha) & \sin(\alpha) \\ -\sin(\alpha) & \cos(\alpha) \end{pmatrix} $. Is there a general formula for
$$B=\begin{pmatrix} 0& \alpha \\ \beta & 0 \end{pmatrix}$$?
Thanks!
| here is another way to find $e^A.$ we will use the interpretation that $x = e^{At}x_0$ is the unique solution of $$\frac{dx}{dt} = Ax, x(0) = x_0.$$
for $$\frac{d}{dt}\pmatrix{x\\y} = \pmatrix{0&1\\-4&0}\pmatrix{x\\y}$$ which in component form is $$\dot x = y, \dot y = -4x \to \ddot x=-4x, \ddot y= -4y \text{ and } x(0) = x_0, y(0) = y_0.$$ the solutions are $$x = x_0\cos 2t+\frac 12y_0 \sin 2t, y = \dot x = -2x_0\sin 2t + y_0\cos 2t$$
writing the last equation in matrix form, we have $$ \pmatrix{x\\y} = \pmatrix{\cos 2t&\frac 12 \sin 2t\\-2\sin 2t&\cos 2t}\pmatrix{x_0\\y_0}.$$ therefore, $$e^{At} = \pmatrix{\cos 2t&\frac 12 \sin 2t\\-2\sin 2t&\cos 2t}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1418210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
} |
If in a triangle $ABC$,$1=2\cos A\cos B\cos C+\cos A\cos B+\cos B\cos C+\cos C\cos A$,then prove that triangle will be equilateral triangle If in a triangle $ABC$ we have
$$1=2\cos A\cos B\cos C+\cos A\cos B+\cos B\cos C+\cos C\cos A\ ,$$
then the triangle will be equilateral triangle.
I tried but except few steps,could not prove it. Putting $A=B=C=\frac{\pi}{3}$ makes both sides equal. How should i prove it?Please help me.
| Observe that the equality fails if one of the $\cos$ is negative. We can assume that triangle is acute. By Jensen and followed by AM-GM inequality:
$x = \cos A, y = \cos B, z = \cos C\to x+y+z \leq \dfrac{3}{2}$:
$1 = 2xyz + xy+yz+zx \leq 2\cdot \dfrac{(x+y+z)^3}{27}+\dfrac{(x+y+z)^2}{3}\leq \dfrac{2}{27}\cdot \left(\dfrac{3}{2}\right)^3+\dfrac{1}{3}\cdot \left(\dfrac{3}{2}\right)^2=1\to $ equality must occur when $x = y = z \to $ triangle is equilateral.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1419328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Calculation of improper integral: $\int_{0}^{\infty}\frac{x^3\ln x}{(x^4+1)^3} \,dx$ One of an exam's task was to calculate the following integral: $$I=\int_{0}^{\infty}\frac{x^3\ln x}{(x^4+1)^3} \,dx$$
I tried integration by parts:
$$I=\frac{1}{4}\int_{0}^{\infty}\ln x \cdot (x^4+1)^{-3} \,d(x^4+1)$$
but then things got messy and I ultimately concluded that the improper integral diverges, which is wrong since the answer is:
$$I=-\frac{1}{32}$$
What should I do?
| $$I = \int\frac{x^3\ln(x)}{(x^4+1)^3}dx=\frac{\ln x}4\int \frac{4x^3}{(x^4+1)^3}dx-\int\left(\frac1{4x}\int \frac{4x^3}{(x^4+1)^3}dx\right)dx $$
$$=-\frac{\ln x}{8(x^4+1)^2}+\frac 1{32}\int \frac{4x^3}{x^4(x^4+1)^2}dx$$
$$=-\frac{\ln x}{8(x^4+1)^2}+\frac1{32}\ln\left(\frac{x^4}{1+x^4}\right)+\frac 1{32(1+x^4)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1422837",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Prove by Induction: $8^n - 3^n$ is divisible by $5$ for all $n \geq 1$ Prove by Induction that for all $n \geq 1$ we have
$$8^n - 3^n \text{is divisible by 5} ...(*)$$
My proof so far
Step 1: For $n=1$ we have $8^1 - 3^1 = 8 - 3 = 5$ which is divisible by 5.
Step 2: Suppose (*) is true for some $n=k\geq 1$ that is $8^k-3^k$ is divisible by 5.
Step 3: Prove that (*) is true for $n=k+1$, that is $8^{k+1} - 3^{k+1}$ is divisible by 5. We have
$$8^{k+1}-3^{k+1} = 8*8^k - 3*3^k$$
Can anyone explain the next logical expansion?
Update:
$$8^{k+1}-3^{k+1} = 8*8^k - 3*3^k = 5*8^k + 3*8^k - 3*3^k = 5*8^k + 3(8^k - 3^k)$$
Now we can say that $5*8^k$ is divisible by 5 since it has the form $5*p$ where $p$ is an integer $\geq 1$
And it is assumed that $8^k-3^k$ is divisible by $5$, then $3(8^k-3^k)$ is divisible by $5$ which means it has the form $5p$
so we reduce the expression to
$$5p+5p = 5(p+p)$$
which is of the form $5p$, which is divisible by $5$.
Is my proof correct?
| Hint: note that $8=5+3$ and then factor.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "9",
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Solve trigonometric inequality $ \sin x+2 \cos x<2$ $$ \sin x+2 \cos x<2$$
$$ \dfrac{2t}{1+t^2}+2\dfrac{1-t^2}{1+t^2}<2$$
$$ 4t^2-2t>0$$
$$ 2t(2t-1)>0$$
$$ t(2t-1)>0$$
$$ (t>0 \wedge t>\dfrac{1}{2}) \vee (t<0 \wedge t<\dfrac{1}{2})$$
From this, I can only find $x<2\pi+2k\pi$, and, $x<2k\pi$, these are good (I think), but I should find another two solutions.
| First, we solve $\sin x + 2 \cos x = 2$
Let $\theta$ be the angle corresponding to the point $(x,y) = (2,1)$ with amplitude $r = \sqrt 5.$ Then
$\cos \theta = \dfrac{2}{\sqrt 5}$ and
$\sin \theta = \dfrac{1}{\sqrt 5}$
Then
\begin{align}
\dfrac{2}{\sqrt 5}\cos x + \dfrac{1}{\sqrt 5} \sin x &= \dfrac{2}{\sqrt 5}\\
\cos x \cos \theta + \sin x \sin \theta &= \dfrac{2}{\sqrt 5}\\
\cos(x - \theta) &= \dfrac{2}{\sqrt 5}\\
\cos(x - \theta) &= \cos \theta\\
x - \theta &= 2n\pi \pm \theta\\
x &\in \{2n\pi + 2\theta : n \in \mathbb Z\} \cup
\{2n\pi : n \in \mathbb Z\}
\end{align}
This splits the unit circle, $x \in [0, 2\pi]$, into four pieces:
$\{0\},\; (0, 2\theta),\; \{2\theta\},\; (2\theta, 2\pi)$
Equality occurs at $x = 0$ and $x = 2\theta$.
$\sin x + 2 \cos x < 2$ will occur in the interval
$(0, 2\theta)$ or the interval
$(2\theta, 2\pi).$
Since $\theta$ is in the first quadrant, $\theta \in (0, 2\theta).$
$\sin \theta + 2 \cos \theta = \sqrt 5 > 2$
So the intervals $(2n\pi, 2n\pi + 2\theta)$ are not part of the solution set.
Since $\theta$ is in the first quadrant, $\pi \in (2\theta, 2\pi).$
$\sin \pi + 2 \cos \pi = -2 < 2$
So the intervals $(2n\pi+2\theta, (2n+1)\pi)$ are part of the solution set.
So the solution set is $x \in (2n\pi+2\theta, (2n+1)\pi)\quad \forall n \in \mathbb Z$ where $\theta = \arcsin \dfrac{1}{\sqrt 5}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1425452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to get formula for sums of powers? Assuming I have Bernoulli numbers:
$B = [\frac{1}{1},\frac{1}{2},\frac{1}{6},\frac{0}{1},-\frac{1}{30}, \frac{0}{1}, \frac{1}{42}, ...]$
How can I get the coefficients of the sums of powers formulas?
For example the sum of squares is $(1/3)n^3 + (1/2)n^2 + 1/6n$
| What you're looking for is Faulhaber's formula (Wikipedia link):
$$\sum_{k=1}^n k^p = {1 \over p+1} \sum_{j=0}^p (-1)^j{p+1 \choose j} B_j n^{p+1-j},\qquad \mbox{where}~B_1 = -\frac{1}{2}$$
expressing the sum of the first $n$ $p$th powers as a polynomial in $n$ whose coefficients involve the Bernoulli numbers and some binomial coefficients.
To get the example you listed,
$$\begin{align*}
\sum_{k=1}^nk^2&=\frac{1}{3}\sum_{j=0}^2(-1)^j\binom{3}{j}B_jn^{3-j}\\\\
&=\left(\frac{1}{3}(-1)^0\binom{3}{0}B_0\right)n^3+\left(\frac{1}{3}(-1)^1\binom{3}{1}B_1\right)n^2+\left(\frac{1}{3}(-1)^2\binom{3}{2}B_2\right)n\\\\
&=\left(\frac{1}{3}\cdot 1\cdot 1\cdot 1\right)n^3+\left(\frac{1}{3}\cdot (-1)\cdot 3\cdot \left(-\frac{1}{2}\right)\right)n^2+\left(\frac{1}{3}\cdot 1\cdot 3\cdot \frac{1}{6}\right)n\\\\
&=\left(\frac{1}{3}\right)n^3+\left(\frac{1}{2}\right)n^2+\left(\frac{1}{6}\right)n
\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1426663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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How to show that $\lim_{x\to 1}(x^3-1)/(x-1)=3$? How to show that
$$
\lim_{x\to 1}\frac{x^3-1}{x-1}=3?
$$
I tried to solve but couldn't. Please help me.
| Let $x = y+1$,
so we want to see what happens as $y \to 0$.
Note:
I almost always want to have
a variable that controls a limit
go to zero
instead of some other value
(like $1$ in this case).
\begin{align}
\frac{x^3-1}{x-1}
&=\frac{(y+1)^3-1}{(y+1)-1}\\
&=\frac{(y^3+3y^2+3y+1)-1}{y}\\
&=\frac{y^3+3y^2+3y}{y}\\
&=y^2+3y+3
\qquad\text{whenever }y \ne 0\\
&\to 3 \qquad\text{as } y \to 0\\
\end{align}
Note that this shows that
$\frac{x^3-1}{x-1}
=(x-1)^2+3(x-1)+3
$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1426908",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Turning 500, 1100, 1800, 2600, 3500, etc into 1, 2, 3, 4, 5, etc I would love to have a formula of the form $n = F(m)$ for this progression ...
n m
---------
1 500
2 1100
3 1800
4 2600
5 3500
...
where the second-differences in $m$ are constant, on up to $n = 100$.
| Divide the elements by $100$ to get the sequence $5,11,18,26,35,...$, which is
1 5
2 5+6
3 5+6+7
4 5+6+7+8
...
which is
1 1+2+3+4+5 - 10
2 1+2+3+4+5+6 - 10
3 1+2+3+4+5+6+7 - 10
...
But these sums are triangle numbers, so the terms are given by $\frac{(n+4)(n+5)}{2} -10$. Now multiply by $100$ to recover the original terms:
$$f(n) = 50(n+4)(n+5)-1000 = 50 n^2 + 450 n$$
which can then be inverted using the quadratic formula to obtain
$$n = \frac{-450 + \sqrt{450^2 + 4\cdot 50\cdot f(n)}} {2\cdot 50}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1428848",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluating $\lim _{x\to 1}\left(\frac{\sqrt[3]{x}-1}{2\sqrt{x}-2}\right)$ I'm trying to evaluate the limit
$$\lim _{x\to 1} \frac{\sqrt[3]{x}-1}{2\sqrt{x}-2} .$$
I used an online limit calculator to find the result, which gives
$$\lim _{x\to 1} \frac{x^{\frac{1}{6}}+1}{2\left(\sqrt[3]{x}+x^{\frac{1}{6}}+1\right)}.$$
Then, plugging the value $1$ for $x$, you get $\frac{1}{3}$.
I don't see how did they reach that conclusion. This is how I tried to tackle it:
$$\frac{\sqrt[3]{x}-1}{2\sqrt{x}-2} = \frac{\sqrt[3]{x}-1}{2\sqrt{x}-2} \cdot \frac{\sqrt[3]{x}+1}{\sqrt[3]{x}+1},$$
which then yields
$$\frac{x-1}{(2\sqrt{x}-2)(\sqrt[3]{x}+1)},$$
and that becomes
$$\frac{x-1}{2\cdot(\sqrt{x}-1)\cdot(\sqrt[3]x+1)}.$$
That's
$$\frac{x-1}{2\cdot(\sqrt[6]{x}+\sqrt{x}-\sqrt[3]{x}-1)},$$
and this will still evaluate to $\frac{0}{0}$.
How did they solve this, exactly?
| Given $\displaystyle \lim_{x\rightarrow 1}\frac{1}{2}\left[\frac{x^{\frac{1}{3}}-1}{x^{\frac{1}{2}}-1}\right] = \frac{1}{2}\times \lim_{x\rightarrow 1}\left[\left(\frac{x^{\frac{1}{3}}-1}{x-1}\right)\times \left(\frac{x-1}{x^{\frac{1}{2}}-1}\right)\right]$
Now Using The formula $\displaystyle \lim_{y\rightarrow 1}\frac{y^n-1}{y-1} = n$
So we get $\displaystyle \frac{1}{2}\times \lim_{x\rightarrow 1}\left[\frac{x^{\frac{1}{3}}-1}{x-1}\right]\times \lim_{x\rightarrow 1}\left[\frac{x-1}{x^{\frac{1}{2}}-1}\right] =\frac{1}{2} \times \frac{1}{3}\times \frac{2}{1} = \frac{1}{3}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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Solve for $x^{11} + \frac{1}{x^{11}}$ if $x^3 + \frac{1}{x^3} = 18.$ Solve for $x^{11} + \frac{1}{x^{11}}$ if $x^3 + \frac{1}{x^3} = 18.$
I'm familiar with variants of this problem, especially those where the exponent in the given is a factor of the exponent in what we want to solve for, but this one stumped me.
| Hint: $$x^3+\frac{1}{x^3}=(x+\frac{1}{x})(x^2+\frac{1}{x^2}-1)=18.$$
Let $$x+\frac{1}{x}=t.$$ Then your equation reduces to $(t(t^2-3))=18$. Solve this cubic and you get $x+\frac{1}{x}.$
Now I think you can carry on after that.
| {
"language": "en",
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Put $(7+5\sqrt{2})^{\frac{1}{3}}$ in the form $x+y(\sqrt{2})$ I said, let:
$(7+5\sqrt{2})^{\frac{1}{3}}=((x+y\sqrt{2})^{3})^{\frac{1}{3}}$
Therefore,
$(7+5\sqrt{2})=(x+y\sqrt{2})^{3}$
Hence,
$(7+5\sqrt{2})=x^{3}+3x^{2}y(\sqrt{2})+3xy^{2}(\sqrt{2})^{2}+y^{3}
(\sqrt{2})^3$
However, from here how do I go? Anyone have any ideas?
Thanks a bunch in advance.
| Hint:
Notice
\begin{align}
7+5\sqrt{2}&=2\sqrt{2}+3(2)(1)+3(\sqrt{2})(1)+1\\
&=(\sqrt{2})^3+3(\sqrt{2})^2(1)+3(\sqrt{2})(1)^2+(1)^3
\end{align}
Can you recognize this pattern?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Help factorising a sixth degree polynomial I have to factorise- $$x^6+5x^3+8$$Answer is $$(x^2−x+2)(x^4+x^3−x^2+2x+4)$$.I have also used factor theorem.Please help me.Thanks in advance.
| Break the equation $x^6+5x^3+8$ into $x^6+8-x^3+6x^3$. It then comes into the form $a^3+b^3+c^3-3abc$. Factorise it using the formula $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$.
| {
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"source": "stackexchange",
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If $(x-a)^2=(x+a)^2$ for all values of $x$, then what is the value of $a$? If $(x-a)^2=(x+a)^2$ for all values of $x$, then what is the value of $a$?
At the end when you get $4ax=0$, can I divide by $4x$ to cancel out $4$ and $x$?
| $$(x-a)^2 = (x+a)^2$$
Expand squares:
$$x^2 - 2ax + a^2 = x^2 + 2ax + a^2$$
Subtract $x^2+a^2$ from both sides:
$$-2ax = 2ax$$
add $2ax$ and divide by $4$:
$$0 = ax$$
So either 1) $a = 0$ or 2) $x = 0$.
*
*$$(x-0)^2 = (x+0)^2 \Leftrightarrow x^2 = x^2$$
Which is true for all x.
*$$(0-a)^2 = (0+a)^2 \Leftrightarrow (-a)^2 = (a)^2$$
Which is true for all $a$.
So either $a=0$ and $x$ can be anything or $x=0$ and $a$ can be anything.
| {
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"source": "stackexchange",
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Prove that $a^2+ab+b^2\ge 0$ How to prove that $a^2+ab+b^2\ge 0$?
Obviously the squares are positive, but how can I be sure that $ab$ doesn't become too negative with a certain combination of $a$ and $b$?
| There is yet another of seeing this.
Since $ab$ is between $-2ab$ and $2ab$, the quantity $a^2+ab+b^2$ must be between $a^2-2ab+b^2$ and $a^2+2ab+b^2$. But these two quantities obey
$$a^2-2ab+b^2=(a-b)^2\ge0 \tag{1}$$
and
$$a^2+2ab+b^2=(a+b)^2\ge0 \tag{2}$$
with equality only if $a=b$ for (1), $a=-b$ for (2),or $a=b=0$ for (1) and (2).
Since $a^2+ab+b^2$ is inclusively between two non-negative quantities, it must be non-negative.
This motivates why the discriminant is so important - in cases such as $a^2+3ab+b^2=(a+b)^2+ab$, the quantity is not between those in (1) and (2), and completing the square will lead to a difference of squares rather than an addition of squares, i.e $(a+\frac{3}{2}b)^2-(\frac{\sqrt5}{2}b)^2$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Maclaurin expansion of $y=\frac{1+x+x^2}{1-x+x^2}$ to $x^4$
Maclaurin expansion of $$\displaystyle y=\frac{1+x+x^2}{1-x+x^2}\,\,\text{to } x^4$$
I have tried by using Maclaurin expansion of $\frac1{1-x}=1+x+x^2+\cdots +x^n+o(x^n)$, but it seems not lead me to anything.
| Hint: notice that
$$y(x) = \frac{1+x}{1-x}\cdot\frac{1-x^3}{1+x^3}\tag{1}$$
and that:
$$ \frac{1+x}{1-x}=1+2x+2x^2+2x^3+\ldots,\tag{2}$$
from which:
$$ \frac{1-x^3}{1+x^3}=1-2x^3+2x^6-2x^9+2x^{12}-\ldots\tag{3}$$
Are you able to multiply $(2)$ and $(3)$ and prove that:
$$ y(x) = 1 + \left(2x+2x^2\right) - \left(2x^4+2x^5\right)+ \left(2x^7+2x^8\right)-\ldots \tag{4}$$
?
| {
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Limit of $f(n)^{g(n)}$ Given the limit:
$$\lim_ {n\to \infty}\left(\frac{n^2+5}{3n^2+1}\right)^{\! n}$$
Is it possible to assume that
$$\lim_ {n\to \infty}\left(\frac{n^2+5}{3n^2+1}\right)^{\! n} = L$$
and then take the natural log of both sides
$$\lim_ {n\to ∞}\left(n \cdot \ln\left(\frac{n^2+5}{3n^2+1}\right)\right) = \ln(L)$$
and keep solving? (I got that $L=0$.)
| First, consider the function
$$\mathrm{f}(x) = \frac{x^2+5}{3x^2+1}$$
The derivative is given by using the quotient rule:
\begin{eqnarray*}
\mathrm{f}'(x) &=& \frac{2x(3x^2+1)-(x^2+5)(6x)}{(3x^2+1)^2} \\ \\
&=& \frac{-28}{(3x+1)^2}
\end{eqnarray*}
This tells us that $\mathrm{f}'(x) < 0$ for all $x \ge 0$, i.e. $\mathrm{f}$ is a decreasing function. This means that $$\mathrm{f}(x) < \mathrm{f}(y) \iff x > y$$
This means that the sequence $\mathrm{f}(n)$, where $n$ is a positive whole number, is strictly decreasing:
$$\mathrm{f}(1) > \mathrm{f}(2) > \mathrm{f}(3) > \ldots > \mathrm{f}(n) > \mathrm{f}(n+1) > \ldots$$
Finally, note that when $n =2$ we have $\mathrm{f}(n) = \frac{9}{13}$, and hence $\mathrm{f}(n) < 1$ for all $n \ge 2$.
Coming back to your question. For all $n \ge 2$, we have
$$0 < \frac{n^2+5}{3n^2+1} < 1 \implies \lim_{n \to \infty}\left( \frac{n^2+5}{3n^2+1} \right)^{\! n} = 0$$
| {
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"source": "stackexchange",
"question_score": "2",
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Integration by partial-fractions, I´m stuck in this one. I don´t really know how get the factors in the denominator which allow me to use a case
$\int\frac{x^2+1}{x^2-x} dx$
| $$\frac{x^2+1}{x^2-x}=\frac{x^2}{x^2-x}+\frac{1}{x^2-x}$$
$$=\frac{x^2}{x(x-1)}+\frac{1}{x(x-1)}$$
$$=\frac{x}{(x-1)}+\frac{1}{x(x-1)}$$
$$=\frac{x-1+1}{(x-1)}+\frac{1}{x(x-1)}$$
$$=1+\frac{1}{(x-1)}+\frac{1}{x(x-1)}$$
now let
$$\frac{1}{x(x-1)}=\frac{A}{x}+\frac{B}{x-1}$$
equate the two sides to find
$$A=-1$$
$$B=1$$
so
$$=1+\frac{1}{(x-1)}+\frac{1}{(x-1)}-\frac{1}{x}$$
$$=1+\frac{2}{(x-1)}-\frac{1}{x}$$
| {
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Prove $2^{2n} = \sum_{k=0}^{n}\binom{2n+1}{k}$ I'm trying to prove the following equation above. So far I have:
\begin{align}
2^{2n} &= (1+1)^{2n}\\
&= \sum_{k=0}^{2n}\binom{2n}{k}1^k1^{n-k} = \sum_{k=0}^{2n}\binom{2n}{k} & \text{(By the Binomial Theorem)}
\end{align}
I know I have to use the following identity somehow:
$$\binom{n+1}{k} = \binom{n}{k-1} + \binom{n}{k}$$
How do I split my summation to get what I'm looking for? Thanks!
EDIT: HERE IS MY SOLUTION
\begin{align*}
2^{2n} &= (1+1)^{2n}\\
&= \sum_{k=0}^{2n}\binom{2n}{k}1^k1^{2n-k} & \text{(By the Binomial Theorem)}\\
&= \sum_{k=0}^{n}\binom{2n}{k} + \sum_{k=n+1}^{2n}\binom{2n}{k}\\
&= \sum_{k=0}^{n}\binom{2n}{k} + \sum_{k=n+1}^{2n}\binom{2n}{2n-k} & \text{(Binomial Symmetry)}\\
&= \sum_{k=0}^{n}\binom{2n}{k} + \sum_{k=0}^{n-1}\binom{2n}{k}\\
&= \sum_{k=0}^{n}\binom{2n}{k} + \sum_{k=1}^{n}\binom{2n}{k-1}\\
&= \binom{2n}{0} + \sum_{k=1}^{n}\binom{2n}{k} + \sum_{k=1}^{n}\binom{2n}{k-1}\\
&= \binom{2n}{0} + \sum_{k=1}^{n}\binom{2n+1}{k} & \text{(By Identity listed above)}\\
&= \sum_{k=0}^{n}\binom{2n+1}{k}
\end{align*}
| I would suggest to use induction:
For $n=1$:
$2^2 = \binom{3}{0}+\binom{3}{1}$, true.
For $m=n+1$:
$2^{2n+2}=4\cdot 2^{2n} = \sum_{k=0}^{n}\binom{2n+1}{k} + \binom{2n+2}{n+1} = 2^{2n} + \binom{2n+2}{n+1} = 2^{2n} + \frac{(2n+2)!}{(n+1)!^2}$
| {
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"timestamp": "2023-03-29T00:00:00",
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How would i go about solving this exponential equation? $$2^{ x-2 }+\frac { 1 }{ \sqrt { { 4 }^{ x+2 } } } =\frac { 17 }{ 16 } $$
I can't seem to find any direction to go that will lead me to a solution. I decided to try to take the natural log all terms in the equation to see where that would lead me. Please do not give me the solution. I just want to be put in the right direction so that I could get the solution on my own. Thanks.
This is what I have done so far:
$$ln(2^{ x-2 })+ln(\frac { 1 }{ \sqrt { { 4 }^{ x+2 } } } )=ln(\frac { 17 }{ 16 } )$$
$$(x-2)ln(2)+[ln(1)-ln(\sqrt { { 4 }^{ x+2 } } )]=ln(17)-ln(16)$$
$$xln(2)-2ln(2)+[ln(1)-ln(4\sqrt { { 4 }^{ x } } )]=ln(17)-ln(16)$$
| There's no need to use logarithms at all here-just indices will do.
Note that: $$\frac{1}{\sqrt{4^{x+2}}}= \frac{1}{2^{x+2}} = \frac{1}{4(2^x)}$$
$$ 2^{x-2} + \frac{1}{2^{x+2}} = \frac{17}{16} $$
$$ \frac{2^x}{4}+ \frac{1}{2^x(4)}= \frac{17}{16}$$
Let $2^x = a$
$$\frac{a}{4}+ \frac{1}{4a}= \frac{17}{16}$$
$$ 4a^2 -17a +4 = 0$$
$$a = 2^{-2}\ \text{or} \ 4 $$
Now substitute the values of $a$ into $2^x = a$ to get the values of $x$.
| {
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a conjectured continued-fraction for $\displaystyle\cot\left(\frac{z\pi}{4z+2n}\right)$ that leads to a new limit for $\pi$ Given a complex number $\begin{aligned}\frac{z}{n}=x+iy\end{aligned}$ and a gamma function $\Gamma(z)$ with $x\gt0$, it is conjectured that the following continued fraction for $\displaystyle\cot\left(\frac{z\pi}{4z+2n}\right)$ is true
$$\begin{aligned}\displaystyle\cot\left(\frac{z\pi}{4z+2n}\right)=\frac{\displaystyle\Gamma\left(\frac{z}{4z+2n}\right)\Gamma\left(\frac{3z+2n}{4z+2n}\right)}{\displaystyle\Gamma\left(\frac{z+n}{4z+2n}\right)\Gamma\left(\frac{3z+n}{4z+2n}\right)}=\cfrac{2z+2n}{2z+n+\cfrac{(0z-n)(4z+3n)} {3(2z+n)+\cfrac{(2z+0n)(6z+4n)}{5(2z+n)+\cfrac{(4z+n)(8z+5n)}{7(2z+n)+\cfrac{(6z+2n)(10z+6n)}{9(2z+n)+\ddots}}}}}\end{aligned}$$
Or in gauss's notation $$\begin{aligned}\displaystyle\cot\left(\frac{z\pi}{4z+2n}\right)=-\frac{1}{2z+2n}\underset{m=0}{\overset{\infty}{\mathbf K}}\frac{((2m-2)z+(m-2)n)((2m+2)z+(m+2)n)}{((2m+1)(2z+n)}\end{aligned}$$
Corollaries:
1):let $z=1$ and $n=2$,then we obtain a beautiful continued fraction for square root 2
$$\begin{aligned}{-1+\cfrac{3}{2+\cfrac{\frac{(-1)(5)}{(1)(3)}} {2+\cfrac{\frac{(1)(7)}{(3)(5)}}{2+\cfrac{\frac{(3)(9)}{(5)(7)}}{2+\cfrac{\frac{(5)(11)}{(7)(9)}}{2+\ddots}}}}}}=\sqrt{2}\end{aligned}$$
2):However the most interesting case(for me at least),occurs when we take the limit to zero
$$\begin{aligned}\lim_{z\to0} \cfrac{z(z+1)}{2z+1+\cfrac{(0z-1)(4z+3)} {3(2z+1)+\cfrac{(2z+0)(6z+4)}{5(2z+1)+\cfrac{(4z+1)(8z+5)}{7(2z+1)+\cfrac{(6z+2)(10z+6)}{9(2z+1)+\ddots}}}}}=\frac{1}{\pi}\end{aligned}$$
yielding a new limit for $\pi$ from which one obtains the first few convergents $\begin{aligned}0,\frac{3}{8},\underline{\frac{5}{16},\frac{15}{47},\frac{7}{22}},\frac{1365}{4288},\frac{3015}{9472},\frac{1575}{4948},\ddots\end{aligned}$.
Where the underlined convergents appear in the stern-brocot tree for $\pi
$ associated to its simple continued fraction.
Q: Is the conjectured continued fraction true (for all complex numbers $z$ with $x\gt0$)?
Update:I initially defined the continued fraction $\displaystyle\cot\left(\frac{z\pi}{4z+2}\right)$ for only natural numbers,but as a matter of fact it holds for all complex numbers $z$ with real part greater than zero.This continued fraction is one special case of the general continued fraction found here.
| O. Perron, Die Lehre von den Kettenbrüchen, Chapter XI, Section 82.
Satz 5 on page 488 (2nd edition, 1927):
Continued fraction
$$
d+\underset{k=1}{\overset{\infty }{\mathbf K}}\;
\frac{a+bk+ck^2}{d+ek}
$$
with $c \ne 0, e \ne 0, e^2+4c \ne 0$ has value
$$
\frac{\sqrt{e^2+4c}\;{}_2F_1(\alpha,\beta;\gamma;x)\;\gamma}{{}_2F_1(\alpha+1,\beta+1;\gamma+1;x)},
$$
where $\alpha,\beta$ are the roots of the quadratic equation
$cZ^2-bZ+a=0$,
$$
\gamma = \frac{b+c}{2c}\left(1-\frac{e}{\sqrt{e^2+4c}\;}\right)
+\frac{d}{\sqrt{e^2+4c}\;}
\\
x=\frac{1}{2}\left(1-\frac{e}{\sqrt{e^2+4c}\;}\right)
$$
Choose the square-root so that
$$
\mathrm{Re}\frac{e}{\sqrt{e^2+4c}\;}>0 .
$$
So substituting in the values in this problem, the question becomes: Is
$$
\cot\left(\frac{z\pi}{4z+2n}\right) =
\frac{\displaystyle(z+n)\sqrt{2}\;
{}_2F_1\left(\frac{-n}{2z+n},\frac{4z+3n}{2z+n};
\frac{3}{2};\frac{1}{2}-\frac{1}{2\sqrt{2}}\right)}
{\displaystyle(2z+n)\;
{}_2F_1\left(\frac{-(2z+2n)}{2z+n},\frac{2z+2n}{2z+n};
\frac{1}{2};
\frac{1}{2}-\frac{1}{2\sqrt{2}}\right)} ?
$$
Writing $z/(4z+2n) = t$ and taking reciprocal, the question becomes: Is
$$
\tan(t\pi) = \frac{{}_2F_1\left(4t-2,-4t+2;
\frac{1}{2};\frac{1}{2}-\frac{1}{2\sqrt{2}}\right)}
{\sqrt{2}(1-2t)\;{}_2F_1\left(4t-1,-4t+3;
\frac{3}{2};\frac{1}{2}-\frac{1}{2\sqrt{2}}\right)} ?
$$
proof for this
Use quadratic transformation
$$
{}_2F_1\left(2a,2b;a+b+\frac{1}{2};u\right)=
{}_2F_1\left(a,b;a+b+\frac{1}{2};4u(1-u)\right)
$$
to get
$$
{}_2F_1\left(4t-2,-4t+2;
\frac{1}{2};\frac{1}{2}-\frac{1}{2\sqrt{2}}\right)
={}_2F_1\left(2t-1,-2t+1;\frac{1}{2};\frac{1}{2}\right) = \sin(\pi t)
$$
$$
{}_2F_1\left(4t-1,-4t+3;
\frac{3}{2};\frac{1}{2}-\frac{1}{2\sqrt{2}}\right)
={}_2F_1\left(2t-\frac{1}{2},-2t+\frac{3}{2};\frac{3}{2};\frac{1}{2}\right)
=\frac{\cos(\pi t)}{\sqrt{2}(1-2t)}
$$
These are known to most CASs.
| {
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iterated sine function on different arguments I want to evaluate the following:
$\lim_{n\rightarrow \infty} \sqrt{n} \sin^{(n)}(2/\sqrt{n})$, where $\sin^{(n)}$ is the iterated sine function.
I do know the proof for $\lim_{n\rightarrow \infty} \sqrt{n} \sin^{(n)}(x_0) =\sqrt{3}$ for any non trivial $x_0 \neq k\pi$ (by using Stolz-Cesaro's theorem with $a_n=\sin^{(n)}(x_0)$,$b_n=n$, and then proving the convergence of $\lim_{n\rightarrow \infty} \frac{1}{a_{n+1}^2}-\frac{1}{a_{n}^2} =1/3$). It seems like the same proof holds for $x_0=1/\sqrt{n}$ and $x_0=2/\sqrt{n}$ and hence proves $\lim_{n\rightarrow \infty} \sqrt{n} \sin^{(n)}(2/\sqrt{n})=\sqrt{3}$. But I am not entirely sure, could you clarify?
|
Claim. For any $x \in \Bbb{R}$ the limit is given by
$$ \lim_{n\to\infty} \sqrt{n} \sin^{\circ n}\left( \frac{x}{\sqrt{n}} \right) = \frac{\sqrt{3}x}{\sqrt{x^2+3}}. $$
Proof. To prove this, WLOG we assume that $x > 0$. Let $f_n(x) = \sqrt{n}\sin(x/\sqrt{n})$ and define
$$ x_{n;0} = x, \qquad x_{n;k+1} = f_n(x_{n;k}). $$
Also, for brevity, we suppress the dependence on $n$ to simply write $x_k = x_{n;k}$ unless we need it explicitly. Now using the inequality $|\sin \theta| \leq |\theta|$, we know that $|x_{k}| \leq |x|$ uniformly in $n$ and $k$. Then it follows that
\begin{align*}
\frac{1}{x_{n}^2}
&= \frac{1}{x^2} + \sum_{k=0}^{n-1} \left( \frac{1}{x_{k+1}^2} - \frac{1}{x_{k}^2} \right) \\
&= \frac{1}{x^2} + \sum_{k=0}^{n-1} \frac{(x_{k} - x_{k+1})(x_{k} + x_{k+1})}{x_{k}^2 x_{k+1}^2} \\
&= \frac{1}{x^2} + \sum_{k=0}^{n-1} \frac{\left( \frac{1}{6n}x_k^3 + \mathcal{O}(n^{-2}x_k^3 ) \right)\left( 2x_k + \mathcal{O}(n^{-1}x_k) \right)}{x_{k}^2 \left( x_k + \mathcal{O}(n^{-1}x_k) \right)^2} \\
&= \frac{1}{x^2} + \frac{1}{3n} \sum_{k=0}^{n-1} \left( 1 + \mathcal{O}(n^{-1} ) \right) \\
&= \frac{1}{x^2} + \frac{1}{3} + \mathcal{O}(n^{-1}).
\end{align*}
It is easy to check that $x_n = x_{n;n}$ is positive when $n$ is large. Consequently we obtain
$$ \lim_{n\to\infty} \sqrt{n} \sin^{\circ n}\left( \frac{x}{\sqrt{n}} \right)
= \lim_{n\to\infty} x_{n;n}
= \frac{\sqrt{3}x}{\sqrt{x^2+3}}. $$
Addendum. The following is a numerical test with $n = 1000$ iterations:
| {
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How do I solve $\lim_{x \to 1} (\frac{1}{x-1}-\frac{2}{x^3-1})$ indeterminate limit without the L'hospital rule? I've been trying to solve this limit without L'Hospital's rule as homework. So I tried rationalizing the denominator and numerator but it didn't work.
My best was: $\lim_{x \to 1} (\frac{1}{x-1}-\frac{2}{x^3-1}) = \lim_{x \to 1} (\frac{1}{x-1}-\frac{2}{(x-1)(x^2+x+1)}) = \lim_{x \to 1} (\frac{x^2+x+1-2}{(x-1)(x^2+x+1)}) = \lim_{x \to 1} (\frac{(x-1)(x+2)}{(x-1)(x^2+x+1)}) = \lim_{x \to 1} (\frac{x+2}{(x^2+x+1)}) = 3 ???$
| this limit doesn't exist since $$\frac{1}{x-1}-\frac{2}{x^3-1}=\frac{x^2+x-1}{(x-1)(x^2+x+1)}$$
| {
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Finding the next limit How to calculate $\displaystyle\lim_{(x,y)\rightarrow(a,a)}\frac{(x-y)a^n+(a-x)y^n-(a-y)x^n}{(x-y)(a-x)(a-y)}$ where $a\neq 0?$
I have the following idea:the factors on the denominator are roots of the expression on numerator because nullify it. So, the polinomyal has these three roots and then, will have a polinomyal expression who has no problem to calculate the limit. Is this idea works?
But if this is true, how can I compute the limit value? An idea to solve this easier?
Thanks in advance.
| In order for the limit to exist it must have the same value along all paths that approach the point $(a,a)$. We will find two paths that disagree for the limit in question.
So first consider the limit along the parametric curve (a line) $x=a-h,y=a+h$, and thus let $h\to0$. We have:
*
*$x-y=-2h$
*$a-y=-h$
*$a-x=h$
So then assuming $n\ge4$
$$\begin{align}
L_1&=\lim_{h\to0}\frac{-2ha^n+h(a+h)^n+h(a-h)^n}{(-2h)(h)(-h)} \\[1em]
&=\lim_{h\to0}\frac{[(a+h)^n-a^n]+[(a-h)^n-a^n]}{2h^2} \\[1em]
\end{align}$$
By the binomial theorem, the numerator is:
$$\begin{align}
N&=[(a^n+na^{n-1}h+\binom{n}{2}a^{n-2}h^2+\binom{n}{3}a^{n-3}h^3)^n+O(h^4)-a^n]\\&+[(a^n-na^{n-1}h+\binom{n}{2}a^{n-2}h^2-\binom{n}{3}a^{n-3}h^3)^n+O(h^4)-a^n] \\
&=2\binom{n}{2}a^{n-2}h^2+O(h^4)
\end{align}$$
So
$$\begin{align}
L_1&=\lim_{h\to0}{\frac{2\binom{n}{2}a^{n-2}h^2+O(h^4)}{2h^2}} \\[1em]
&=\lim_{h\to0}\left\{\binom{n}{2}a^{n-2}+O(h^2)\right\} \\[1em]
&=\binom{n}{2}a^{n-2} \\[1em]
\end{align}$$
This also holds for $2\le n<4$ as higher order terms fall away to zero.
Now consider the limit along the line $y=x$. Then
$$\require{cancel}\begin{align}
L_2&=\lim_{x=y\to a}\frac{(x-y)a^n+(a-x)y^n-(a-y)x^n}{(x-y)(a-x)(a-y)} \\[1em]
&=\lim_{x=y\to a}\frac{(x-y)a^n+(x-a)(x^n-y^n)}{(x-y)(a-x)(a-y)} \\ \end{align}$$
For $n\ge2$, the factorisation $x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+\cdots+xy^{n-2}+y^{n-1})$
means that there is a factor of $(x-y)$ in the numerator and the denominator, so
$$\begin{align}
L_2&=\lim_{x=y\to a}\frac{a^n+(x-a)(x^{n-1}+x^{n-2}y+\cdots+xy^{n-2}+y^{n-1})}{(a-x)^2} \\[1em]
&=\lim_{x\to a}{\frac{a^n}{(a-x)^2}}\qquad(a^n\text{ dominates the second term in the numerator}) \\[1em]
&=\begin{cases}+\infty,&\text{if }a>0 \\-\infty,&\text{if }a<0 \end{cases}
\end{align}$$
The fact that $L_1\ne L_2$ is enough to show that the original limit does not exist, not even in the sense that it is infinite.
Addendum: $n=1$
$$\begin{align}
L&=\lim_{(x,y)\to(a,a)}{\frac{(x-y)a+(a-x)y-(a-y)x}{(x-y)(a-x)(a-y)}} \\[1em]
&=\lim_{(x,y)\to(a,a)}{\frac{0}{(x-y)(a-x)(a-y)}} \\[1em]
\end{align}$$
which is (again) indeterminate if $(a,a)$ is approached along the line $y=x$.
| {
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"url": "https://math.stackexchange.com/questions/1450024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is -2 is a root of the equation : $\sqrt {x^2 - 8} = \sqrt {3x + 2}$? Is -2 is a root of the equation : $\sqrt{x^2 - 8} = \sqrt{3x + 2}$ ?
Is it any limitation of the root of this equation?
If the root is -2, both sides of the equation are equal to 2i.
Is this acceptable since the variable x has no limitation.
| Notice, the given equation is $\sqrt{x^2-8}=\sqrt{3x+2}$, now squaring both the sides we get $$x^2-8=3x+2$$ $$ x^2-3x-10=0$$ $$(x+2)(x-5)=0$$ On solving we get $$x=-2, \ 5$$
Now, let's check out both the roots, if they are acceptable by satisfying the given (original) equation
Substituting $x=-2$, we get $LHS=RHS=\sqrt{-2}$ which is undefined value hence, the $x=-2$ is unacceptable i.e. it's not a root of the given equation.
While, Substituting $x=5$, we get $LHS=RHS=\sqrt{17}$ which is defined value hence, the $x=5$ is a root of given equation it is not $x=-2$
| {
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"url": "https://math.stackexchange.com/questions/1451868",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Another beautiful arctan integral $\int_{1/2}^1 \frac{\arctan\left(\frac{1-x^2}{7 x^2+10x+7}\right)}{1-x^2} \, dx$ Do you think we can express the closed form of the integral below in a very nice and short way?
As you already know, your opinions weighs much to me, so I need them!
Calculate in closed-form
$$\int_{1/2}^1 \frac{\arctan\left(\frac{1-x^2}{7 x^2+10x+7}\right)}{1-x^2} \, dx.$$
I'm looking forward to your precious feedback!
Mathematica tells us the closed form is
$$\frac{1}{4} i \text{Li}_2\left(\frac{1}{5}+\frac{2 i}{5}\right)+\frac{1}{4} i \text{Li}_2\left(\frac{2}{15}+\frac{2 i}{5}\right)-\frac{1}{4} i \text{Li}_2\left(\frac{1}{5}-\frac{2 i}{5}\right)-\frac{1}{4} i \text{Li}_2\left(\frac{2}{15}-\frac{2 i}{5}\right)-\frac{1}{4} i \text{Li}_2\left(\frac{1}{4}+\frac{i}{8}\right)+\frac{1}{4} i \text{Li}_2\left(\frac{1}{4}-\frac{i}{8}\right)-\frac{1}{4} i \text{Li}_2\left(\frac{1}{10}+\frac{3 i}{10}\right)+\frac{1}{4} i \text{Li}_2\left(\frac{1}{10}-\frac{3 i}{10}\right)+\frac{1}{4} i \text{Li}_2\left(\frac{1}{4}+\frac{i}{12}\right)-\frac{1}{4} i \text{Li}_2\left(\frac{1}{4}-\frac{i}{12}\right)-\frac{1}{4} i \text{Li}_2\left(\frac{4}{15}+\frac{8 i}{15}\right)+\frac{1}{4} i \text{Li}_2\left(\frac{4}{15}-\frac{8 i}{15}\right)+\frac{1}{4} \log (4) \tan ^{-1}(6)-\frac{1}{4} \log (4) \tan ^{-1}(9)+\frac{1}{4} \log (4) \tan ^{-1}\left(\frac{1}{6}\right)-\frac{1}{4} \log (4) \tan ^{-1}\left(\frac{1}{9}\right)-\frac{1}{4} \log (9) \tan ^{-1}(2)+\frac{1}{4} \log (9) \tan ^{-1}(3)-\frac{1}{4} \log (9) \tan ^{-1}(6)+\frac{1}{4} \log (9) \tan ^{-1}(9)-\frac{1}{4} \log (9) \tan ^{-1}\left(\frac{3}{55}\right)+\frac{1}{4} \log (16) \tan ^{-1}(2)-\frac{1}{4} \log (16) \tan ^{-1}(3).$$
| It can be easily checked by differentiation that:
$${\large\int}\,\frac{\arctan\left(\frac{1-x^2}{7 x^2+10x+7}\right)}{1-x^2} \, dx=\\
\frac i4\left[\vphantom{\Large|}\operatorname{Li}_2\left(\left(-\tfrac12+\tfrac i6\right)(x-1)\right)-\operatorname{Li}_2\left(\left(-\tfrac12-\tfrac i6\right)(x-1)\right)\\
\,+\operatorname{Li}_2\left(\left(-\tfrac12-\tfrac i4\right)(x-1)\right)-\operatorname{Li}_2\left(\left(-\tfrac12+\tfrac i4\right)(x-1)\right)\\
\!\!\!\!\!\!+\operatorname{Li}_2\left(\left(\tfrac12+i\right)(x+1)\right)\,\,\,\,\,\,-\operatorname{Li}_2\left(\left(\tfrac12-i\right)(x+1)\right)\\
\,\,\,\,\,\,\,\,\,\,+\operatorname{Li}_2\left(\left(\tfrac12-\tfrac{3i}2\right)(x+1)\right)\,\,\,-\operatorname{Li}_2\left(\left(\tfrac12+\tfrac{3i}2\right) (x+1)\right)\right]\color{gray}{+C}$$
This enables us to evaluate a definite integral over any interval.
If you are only interested in real values of $x$ then the antiderivative can be simplified to:
$$\frac12\Im\left[\vphantom{\Large|}\operatorname{Li}_2\left(\left(\tfrac12+\tfrac{3i}2\right)(x+1)\right)-\operatorname{Li}_2\left(\left(\tfrac12+i\right)(x+1)\right)\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\operatorname{Li}_2\left(\left(\tfrac i4-\tfrac12\right)(x-1)\right)-\operatorname{Li}_2\left(\left(\tfrac i6-\tfrac12\right)(x-1)\right)\right]\color{gray}{+C}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1453345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 2
} |
Proving tan((x + y)/2) = (sin x + sin y)/(cos x + cos y) with the angle sum and difference identities I've recently come across this problem of proving
$$ \tan \frac{x + y}{2} = \frac{\sin x + \sin y}{\cos x + \cos y} $$
Not a difficult problem, I thought. I would have rewritten the RHS using the sum-to-product identities of sine and cosine.
But the solution given in the back of the book is
\begin{align}
\tan \frac{x + y}{2} &= \frac{\sin x \sin y}{\cos x + \cos y} \\
&= \frac{2\sin(\frac{x}{2})\cos(\frac{x}{2})+2\sin(\frac{y}{2})\cos(\frac{y}{2})}{2\cos^2(\frac{x}{2})-1 + 1 - 2\sin^2(\frac{y}{2})} \\
&= \frac{\sin(\frac{x}{2})\cos(\frac{x}{2}) + \sin(\frac{y}{2})\cos(\frac{y}{2})}{\cos^2(\frac{x}{2}) - \sin^2(\frac{y}{2})} \tag{1} \\
&= \frac{\sin(\frac{x+y}{2})}{\cos(\frac{x+y}{2})} \tag{2} \\
&= \tan \frac{x + y}{2}
\end{align}
What I'm confused about is the step from (1) to (2). It appears they tried to use the angle sum identities, but Wikipedia says that the formulas are actually
$$ \sin(\alpha \pm \beta) = \sin \alpha \cos \beta \pm \cos \alpha \sin \beta \\
\cos(\alpha \pm \beta) = \cos \alpha \cos \beta \mp \sin \alpha \sin \beta
$$
which don't actually match the numerator or denominator in (1). Can the angle sum identities be used like this?
I'm also interested in knowing how one would actually go from (1) to (2) without the angle sum identities. It seems like there would be a few intermediate steps.
| \begin{align}
\frac{\sin x + \sin y}{\cos x + \cos y}
&= \frac{\sin\left(\frac{x+y}{2} + \frac{x-y}{2}\right)
+ \sin\left(\frac{x+y}{2} - \frac{x-y}{2}\right)}
{\cos\left(\frac{x+y}{2} + \frac{x-y}{2}\right)
+ \cos\left(\frac{x+y}{2} - \frac{x-y}{2}\right)}\\
&= \frac{2\sin\left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right)}
{2\cos\left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right)}\\
&= \tan\left(\frac{x+y}{2}\right)
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1453588",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
inverse of $\arcsin (\frac{x}{x-1})$ determine the inverse of
a) $y=\arcsin \left(\dfrac{x}{x-1}\right)$
b) $y= \dfrac{1-2e^{-x}}{4}$
I learned you the steps for finding the inverse are
1) get it in a form of $x= \dots$
2) change $x$ and $y$
a) $y=\arcsin \left(\dfrac{x}{x-1}\right)$
is $y=\arcsin \left(\dfrac{x}{x-1}\right)\quad \iff \quad \sin y = \dfrac{x}{x-1}$? and if so what next?
b) $y= \dfrac{1-2e^{-x}}{4} \quad \iff \quad 4y-1=-2e^{-x} \quad\iff\quad -2y + \frac{1}{2}= e^{-x}$ . but then I'm stuck.
I would really appreciate the help!
| $\sin y(x-1)=x\Rightarrow \sin y x-x=siny \Rightarrow x(\sin y -1)=\sin y$ so $$x=\frac{\sin y}{\sin y -1}\Rightarrow y^{-1}=\frac{\sin x}{\sin x -1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1454218",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
isomorphic equations W. Ross Ashby states in his book "An Introduction to Cybernetics" that the system:
$\ x' = 1/2(x^2+y^2) + x*y + y$
$\ y' = 1/2(x^2+y^2) + x*y + x$
is isomorphic to the system:
$\ u' = -u$
$\ v' = v + v^2$
under the transformation:
$\ u = x - y$
$\ v = x + y$
can someone explain this properly?
| Note that you can express $x$ and $y$ in terms of $u$ and $v$:
$$x=\frac{u+v}2\\y=\frac{-u+v}2$$
So, from the first equation of your system,
$$x' = \frac12(x^2+y^2) + xy + y,$$
we have
$$\left(\frac{u+v}2\right)' = \frac12\left(\left(\frac{u+v}2\right)^2+\left(\frac{-u+v}2\right)^2\right) + \left(\frac{u+v}2\right)\left(\frac{-u+v}2\right) + \left(\frac{-u+v}2\right)\\
\frac{u'+v'}2 = \frac12\left(\frac{u^2+v^2}{2}\right) + \frac{-u^2+v^2}4\ + \frac{-u+v}2\\
\frac{u'+v'}2 = \frac{-u+v+v^2}2$$
$$
u'+v' = -u+v+v^2\tag{$*$}
$$
So that simplified nicely. Now do the same with the second equation, put the result together with $(*)$, and see what happens.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1456637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What is the following limit? $\lim\limits_{x\to 0}=\frac{(1+x)^5-1-5x}{x^2+x^5}$ What is the following limit?
$$\lim\limits_{x\to 0}=\frac{(1+x)^5-1-5x}{x^2+x^5}$$ Should I calculate the exact value of $(1+x)^5$?
| It suffices to know the coefficient of $x^2$ in the numerator.
Note that $(1+x)^5-1-5x=\binom{5}{2}x^2+\mathcal{O}(x^3)$ and $x^2+x^5=x^2+\mathcal{O}(x^3)$ and hence the limit will be
$$\lim_{x \to 0} \frac{10x^2+\mathcal{O}(x^3)}{x^2+\mathcal{O}(x^3)}=10$$
Alternatively, if you don't want to use these symbols, just cancel out $x^2$ and obtain
$$\lim_{x \to 0} \frac{10+ax+bx^2+cx^3}{1+x^3}=10$$
Of course, it is not hard to compute the values of $a,b,c$ using binomial coefficients but it is not necessary for determining the limit.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1458378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve absolute value inequality I have to show the inequality
$$
\left|\frac{1}{2 + a}\right| < 1.
$$
How do I do this?
I know that a fraction is less than 1 when the denominator is greater than the numerator, but I cannot just check if $2 + a > 1$ because of the absolute value sign.
Edit
If I use
$$
\left|\frac{1}{2 + a}\right| < 1 \Leftrightarrow -1 < \frac{1}{2 + a} < 1.
$$
I have to look at the inequalities separately, i.e. $\frac{1}{2+a} > -1 \Leftrightarrow 1 > -2 - a \Leftrightarrow a > -3$ and $\frac{1}{2+a} < 1 \Leftrightarrow 1 < 2+a \Leftrightarrow a > -1$.
Since $a > -1 > -3$, $a$ must just be greater than $-1$. But what about $a = -2$ which yields a zero in the denominator?
| Both sides are positive, so you can take their reciprocals (of course the 'less than' flips to 'greater than'):
$$\left|\frac 1{2+a}\right| < 1 \iff \frac 1{\left|\frac 1{2+a}\right|} = \frac {|2+a|}{|1|}= |2+a| > 1$$
That is equivalent to an alternative:
$$(2+a) < -1 \lor (2+a) > 1$$
which resolves to:
$$a < -3 \lor a > -1$$
Equivalently
$$a\in (-\infty, -3)\cup (-1,\infty)$$
EDIT in reply to the comment
No, $1/(2+a)>−1$ does not imply $a>−3$.
When you multiply both sides by $(2+a)$ you must consider the sign of the multiplicand term. If the term is negative, the direction of an inequality gets reversed. So you have two possible cases here:
$$\color{red}{1/(2+a) > -1} \quad |\,\times(2+a)$$
$$\begin{cases}1 > -1\times(2+a) & \text{ if}\ (2+a) > 0 \\
\qquad \text{or} \\
1 < -1\times(2+a) & \text{ if}\ (2+a) < 0 \end{cases}$$
This is equivalent to
$$1 > -2-a \ \text{and}\ 2+a > 0 \ \text{or} \ 1 < -2-a\ \text{and}\ 2+a < 0$$
$$a > -3 \ \text{and}\ a > -2 \ \text{or} \ a < -3\ \text{and}\ a <-2$$
Finally
$$\color{red}{a > -2 \ \text{or} \ a < -3}$$
Similary from the other inequality we get
$$\color{green}{1/(2+a)<1} \quad |\,\times(2+a)$$
$$\begin{cases}1 < 1\times(2+a) & \text{ if}\ (2+a) > 0 \\
\qquad \text{or} \\
1 > 1\times(2+a) & \text{ if}\ (2+a) < 0 \end{cases}$$
$$1 < 2+a \ \text{and}\ 2+a > 0 \ \text{or} \ 1 > 2+a\ \text{and}\ 2+a < 0$$
$$a > -1 \ \text{and}\ a > -2 \ \text{or} \ a < -1\ \text{and}\ a <-2$$
$$\color{green}{a > -1 \ \text{or} \ a < -2}$$
Together they make
$$(\color{red}{a > -2 \ \text{or} \ a < -3}) \ \text{and} \ (\color{green}{a > -1 \ \text{or} \ a < -2})$$
so
$$a < -3 \ \text{or} \ a > -1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1460684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proving a property of a prime number (Elementary number theory) Prove the following statement.
For all prime numbers $a$, $b$, and $c$, $a
^2 + b^2
\neq c^2$
.
What i tried
Proving by contradiction
Assume the negation of the statement
There exists prime numbers $a$, $b$, and $c$ such that $a
^2 + b^2
= c^2$
Since we know that all prime number except $2$ is also an odd number
Then we let $a=2n+1$ $b=2l+1$
Then we have
$4l^{2}+4l+1+4k^{2}+4k+1=c^{2}$
$4l^{2}+4l+4k^{2}+4k+2=c^{2}$
$2(2l^{2}+2l+2k^{2}+2k+1)=c^{2}$
From this expression,since $c^{2}$ is even then $c$ is even. But we have a case where $c=2$, so if $c=2$, then $c^{2}=4$ and
$2l^{2}+2l+2k^{2}+2k+1=2$. Then
$2l^{2}+2l+2k^{2}+2k=1$ and
$l^{2}+l+k^{2}+k=0.5$
Which is not possible since the sum of integers have to also be an integer
Which contradicts with the face that $c$ is either odd or has a value of $2$ for
it to be a prime number
Is my proof correct.Could anyone explain. Thanks
| We cannot find primes $a,b,c$ such that $a^{2} + b^{2} = c^{2}$ because of an elementary classical result: Take any integers $a,b,c$ such that $(a,b) = 1$; then $a^{2} + b^{2} = c^{2}$ if and only if $a = u^{2} - v^{2}$, $b = 2uv$, $c = u^{2} + v^{2}$ for some integers $u, v$ such that $(u,v) = 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1461241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
} |
How to evaluate the limit of $\frac {2\cos({x-1})-2}{({x²}-2\sqrt{x}+1 )}$ when $x\to1$ without using L'Hospital's rule?
How do I evaluate this without using L'Hospital's rule:$$\lim_{x \to 1}\frac {2\cos({x-1})-2}{({x²}-2\sqrt{x}+1 )}\ ?$$
Note : I used L'Hospital's Rule I find $\frac{-4}{3}$ but in wolfram alpha is $0$
| Set $x-1=u$ and rewrite the expression:
*
*$2\cos(x-1)-2=2\cos u-2=-u^2+o(u^2)$,
*$\begin{aligned}[t]x^2-2\sqrt x+1&=(1+u)^2-2\sqrt{1+u}+1=1+2u+u^2-2-u+\dfrac{u^2}4+o(u^2)+1\\&=u+\dfrac{5u^2}4+o(u^2)
\end{aligned}.$
Thus $2\cos(x-1)-2\sim_1 -(x-1)^2$, $\;x^2-2\sqrt x+1\sim_1 x-1$, and
$$\dfrac{2\cos(x-1)-2}{x^2-2\sqrt x+1}\sim_1\frac{-(x-1)^2}{x-1}=1-x \xrightarrow[x\to 1]{} 0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1461676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
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Showing $\pi/(2\sqrt3)=1-1/5+1/7-1/11+1/13-1/17+1/19-\cdots$ I am struggling to show that $$\dfrac \pi{2\sqrt3}=1-\dfrac 15+\dfrac 17-\dfrac 1{11}+\dfrac 1{13}-\dfrac 1{17}+\dfrac 1{19}-\cdots$$ by using the Fourier series $$\frac \pi2-\frac x2=\sum_1^\infty \dfrac {\sin(nx)}{n}.$$
Can somebody give me any hint?
| This sum may be tackled using the residue theorem.
$$S=\sum_{k=0}^{\infty} \left ( \frac1{6 k+1} - \frac1{6 k+5}\right ) = \frac12 \sum_{k=-\infty}^{\infty} \left ( \frac1{6 k+1} - \frac1{6 k+5}\right )$$
Thus, consider the contour integral
$$\oint_{C_N} dz \frac{2 \pi \, \cot{\pi z}}{36 \left (z+\frac16\right ) \left (z+\frac{5}{6}\right )} $$
where $C_N$ is the square with vertices at $z =\pm (N \pm 1/2)$. As $N \to \infty$, the integral is zero. Thus the sum of the residues at the poles $z=k, k \in \mathbb{Z}$ and $z=-1/6$ and $z=-5/6$ is zero. We thus have
$$\sum_{k=0}^{\infty} \left ( \frac1{6 k+1} - \frac1{6 k+5}\right ) =-\frac{\pi}{18} \left [\frac{\cot{\left ( -\frac{\pi}{6} \right )}}{\frac{4}{6}} + \frac{\cot{\left ( -\frac{5 \pi}{6} \right )}}{-\frac{4}{6}} \right ] = 2 \sqrt{3} \frac{\pi}{18} \cdot \frac{3}{2} = \frac{\pi}{2 \sqrt{3}}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Difficult integral evaluation I'm working through Vector Calculus by Marsden and Tromba to review for my GRE (and because it has really interesting historical snippets) and I ran into a wall on a problem where I have to evaluate the arc length of the vector valued function $<t, tsint, tcost>$ for $0 \leq t \leq \pi$. Here is what I have so far:
$$\int_{0}^{\pi} \sqrt{1^2 + (\sin(t) + t\cos(t))^2 + (\cos(t) - t \sin(t))^2} dt\rightarrow\int_{0}^{\pi} \sqrt{1 + (\sin(t))^2 + (\cos(t))^2 + t^2((\cos(t))^2 + (\sin(t))^2)} dt\\\rightarrow \int_{0}^{\pi} \sqrt{2+t^2} dt$$
But once I get there I cant figure out how to evaluate the integral. I plugged it into wolfram alpha in hopes that by seeing what they get I would have some insight on how to proceed but I have no idea (they had a term with $\sinh^{-1}(t)$).
| You can use integration by parts, or you can use the hyperbolic substitution $t = \sqrt{2}\sinh(u)$. I'll use the latter.
With $t = \sqrt{2}\sinh(u)$, we have $\sqrt{2 + t^2} = \sqrt{2}\cosh(u)$ and $dt = \sqrt{2}\cosh(u)\, du$, so
$$\int_0^\pi \sqrt{2+t^2}\, dt = \int_0^{\sin^{-1}(\pi/\sqrt{2})} 2\cosh^2(u)\, du.$$
Using the identity $2\cosh^2(u) = 1 + \cosh(2u)$, we get
\begin{align}\int_0^{\sinh^{-1}(\pi/\sqrt{2})} 2\cosh^2(u)\, du &= \int_0^{\sinh^{-1}(\pi/\sqrt{2})} (1 + \cosh(2u))\, du \\
&= u + \frac{\sinh(2u)}{2}\bigg|_{u = 0}^{u = \sinh^{-1}(\pi/\sqrt{2})}\\
& = \sinh^{-1}\left(\frac{\pi}{\sqrt{2}}\right)+ \frac{\sinh(2\sinh^{-1}(\pi/\sqrt{2}))}{2}.
\end{align}
From the identity $\sinh(2x) = 2\sinh(x)\, \cosh(x)$ and the computations $\sinh(\sinh^{-1}(\pi/\sqrt{2})) = \pi/\sqrt{2}$ and $\cosh(\sinh^{-1}(\pi/\sqrt{2})) = \sqrt{1 + (\pi/\sqrt{2})^2} = \sqrt{(2 + \pi^2)/2}$, we find
$$\sinh(2\sinh^{-1}(\pi/\sqrt{2})) =2\cdot \frac{\pi}{\sqrt{2}}\cdot \sqrt{\frac{2 + \pi^2}{2}} = \pi\sqrt{2 + \pi^2}.$$
Hence,
$$\int_0^{\pi} \sqrt{2 + t^2}\, dt = \sinh^{-1}\left(\frac{\pi}{\sqrt{2}}\right) + \frac{1}{2}\pi\sqrt{2 + \pi^2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1464215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
$x^{2000} + \frac{1}{x^{2000}}$ in terms of $x + \frac 1x$. If $x + \frac{1}{x} = 1$, then what is
$$ x^{2000} + \frac{1}{x^{2000}} = ?$$
| $$x^2-x+1=0\implies x^3+1=(x+1)(x^2-x+1)=0\iff x^3=-1$$
$$x^{2000}=(x^3)^{666}\cdot x^2=(-1)^{666}\cdot x^2=x^2$$
Finally use $a^2+b^2=(a+b)^2-2ab$ for $x^2+\dfrac1{x^2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1465320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How does $\tan^{-1}(x-\sqrt{1+x^2})=\frac{1}{2}\tan^{-1}x+C$ directly? I'm teaching baby calculus recitation this semester, and I meet a problem to calculate the derivative of
$$y=\tan^{-1}(x-\sqrt{1+x^2})$$
Just apply the chain rule and after some preliminary algebra, I find
$$\frac{dy}{dx}=\frac{1}{2(1+x^2)}$$
What surprises me is that the result implies
$$y=\frac{1}{2}\tan^{-1}x+C$$
Can anyone tell me how to see that directly?
| First let's figure out what $C$ is. Letting $x=0$, we see that $\arctan(-1) = C$ (i.e. $C = -\frac{\pi}{4}$ or $C = \frac{3\pi}{4}$). Since inverse trigonometric functions are hard to work with, let's take $\tan$ of both sides. Doing so we get
$$\tan\left(\arctan(x-\sqrt{1+x^2})\right) = \tan\left(\frac{1}{2}\arctan(x)+C\right).$$
The left hand side is nothing more than $x-\sqrt{1+x^2}$. As for the right hand side, we can employ the tangent sum formula:
$$\tan(A+B) = \frac{\tan A + \tan B}{1-\tan A\tan B}.$$
Thus
$$\tan\left(\frac{1}{2}\arctan(x)+C\right) = \frac{\tan\left(\frac{1}{2}\arctan(x)\right) + \tan(\arctan(-1))}{1-\tan\left(\frac{1}{2}\arctan(x)\right)\tan(\arctan(-1))}.$$
We have that $\tan\left(\frac{1}{2}\arctan(x)\right) = \frac{\sqrt{1+x^2}-1}{x}$ which can be verified by half angle formulas. If you put these pieces together and do some algebra, you will have your answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1466415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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I want know if my logic in showing that $\Bbb Q(\sqrt 2,\sqrt 3) = \Bbb Q (\sqrt 2 + \sqrt 3)$ is correct I want to know if my logic in showing that $\Bbb Q(\sqrt 2,\sqrt 3) = \Bbb Q (\sqrt 2 + \sqrt 3)$ is correct
Now firstly, it seems that $\Bbb Q(\sqrt 2,\sqrt 3)=(\Bbb Q (\sqrt 2))(\sqrt3){\overset{{\huge\color\red?}}{=}}\{a+b\sqrt2+c\sqrt 3:a,b,c\in \Bbb Q\}$
But from my last question it seems that we need to check this for multiplicative closure, $(a+b\sqrt2 + c\sqrt3)(d+e\sqrt 2 + f\sqrt 3)=ad+ae\sqrt 2+ af\sqrt 3 + 2eb+bf\sqrt 3+cd\sqrt 3+ ce\sqrt2\sqrt3+ 3cf$
So we actually find I believe that:
$\Bbb Q(\sqrt2,\sqrt3)=\{a+b\sqrt 2+c\sqrt 3+ d\sqrt2\sqrt3:a,b,c,d\in \Bbb Q\}$?
I believe that $\Bbb Q(\sqrt 2 +\sqrt 3)\overset{\huge{\color\red ?}}=\{a+b(\sqrt 2+ \sqrt 3):a,b\in \Bbb Q\}$
So let's verify multiplicative closure $(a+b\sqrt 2+ b\sqrt 3)(c+d\sqrt 2+ d\sqrt 3)$
$$=ac+ad\sqrt 2+ad\sqrt 3+bc\sqrt2+2bd+bd\sqrt3+bc\sqrt3+bd\sqrt2\sqrt3+3bd$$
$$=(ac+3bd+2bd)+(ad+bc)\sqrt 2 + (ad+bd+bc)\sqrt 3 +(bd)\sqrt2\sqrt3$$
Since $\sqrt2$ and $\sqrt 3$ and $\sqrt2\sqrt3$ don't share common coefficients, they are linearly independent and hence $\Bbb Q(\sqrt2+\sqrt 3)=\Bbb Q(\sqrt2,\sqrt3)$(from above, i.e we have deduced that:) $\Bbb Q(\sqrt{2}+\sqrt{3})=\{a+b\sqrt2+c\sqrt3:a,b,c\in\Bbb Q\}$.
Is that the correct way to show this? Also this means we have:
$$\begin{matrix}&&\left(\Bbb Q(\sqrt{2}+\sqrt{3})=\Bbb Q(\sqrt2,\sqrt3)\right)\\\\&{\huge\diagup}&&{\huge\diagdown}\\\Bbb Q(\sqrt2)&&&&\Bbb Q(\sqrt 3)\end{matrix}$$
| The "usual" proof that $\Bbb Q(\sqrt{2},\sqrt{3}) \subseteq \Bbb Q(\sqrt{2}+ \sqrt{3})$ ), the other inclusion is obvious.
It suffices to show that $\sqrt{2} \in \Bbb Q(\sqrt{2}+ \sqrt{3})$, since then $\sqrt{3} = (\sqrt{2} + \sqrt{3}) - \sqrt{2}$ must be as well.
Now $(\sqrt{2} + \sqrt{3})^3 = 2\sqrt{2} + 6\sqrt{3} + 9\sqrt{2} + 3\sqrt{3}$
$= 11\sqrt{2} + 9\sqrt{3}$.
Hence $\dfrac{1}{2}[(\sqrt{2} + \sqrt{3})^3 - 9(\sqrt{2} + \sqrt{3})] = \sqrt{2}$, QED.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1466935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Line L intersects perpendicularly both the lines $\frac{x+2}{2}=\frac{y+6}{3}=\frac{z-34}{-10}$ and $\frac{x+6}{4}=\frac{y-7}{-3}=\frac{z-7}{-2}$ A straight line L intersects perpendicularly both the lines $\frac{x+2}{2}=\frac{y+6}{3}=\frac{z-34}{-10}$ and $\frac{x+6}{4}=\frac{y-7}{-3}=\frac{z-7}{-2}$,then square of the perpendicular distance of origin from L is ?
Let $\frac{x-x_1}{l}=\frac{y-y_1}{m}=\frac{z-z_1}{n}$ be the required line.
Then $2l+3m-10n=0,4l-3m-2n=0$
So $l:m:n=1:1:\frac{1}{2}$
So the line is $\frac{x-x_1}{1}=\frac{y-y_1}{1}=\frac{z-z_1}{\frac{1}{2}}$.
But i could not solve further.Please help me.
| Broad Steps
(1) We have the dr's of the two given lines
(2) Let P & Q be points on the two lines so that PQ is the shortest distance.
Find equation of L through PQ
(3) Find distance of L from origin.
Details
Note that $\color{blue}{(2,3,-10)}$ and $\color{blue}{(4,-3,-2)}$ are direction ratios of the 2 given lines.
Let $P(2r_1-2, 3r_1 -6, -10r_1 +34)$ and $Q(4r_2-6, -3r_2 + 7, -2r_2 +7)$ be the points on the given lines so that $PQ$ is the line of shortest distance between the two lines.
The direction ratio's of $PQ$ are $(2r_1 - 4r_2 +4, 3r_1 + 3r_2 - 13, -10r_1 + 2r_2 +27)$
Since it is perpendicular to the 2 given lines, we will have
$$\color{blue}{2}(2r_1 - 4r_2 +4) + \color{blue}{3}(3r_1 + 3r_2 - 13) \color{blue}{-10}(-10r_1 + 2r_2 +27) = 0$$ and
$$\color{blue}{4}(2r_1 - 4r_2 +4) \color{blue}{-3(}3r_1 + 3r_2 - 13) \color{blue}{-2}(-10r_1 + 2r_2 +27) = 0$$
These simplify to $113r_1 - 19r_2 - 301 = 0$ and $19r_1 - 29r_2 + 1 = 0$ which gives $\color{blue}{r_1 = 3, r_2 = 2}$
So we now have $P(4,3,4)$ and $Q(2,1,3)$ and direction ratios of $PQ$ as $2,2,1$
Finally, the equation of the line through $PQ$, or line $L$ is
$$\color{blue}{\frac{x-4}{2} = \frac{y-3}{2} = \frac{z-4}{1}}$$
The square of the distance of this line from the origin can be deduced using standard formulae.
However, in order to complete the solution, we can do the following: $(2r +4, 2r + 3, r +4)$ is any point on this line $L$. The square of its distance from the origin is clearly $(2r+4)^2 + (2r+3)^2 + (r+4)^2 = 9(r+2)^2 + 5$ which is minimum when $r = -2$ and the square of the distance is then $\color{blue}{5}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1470121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the area of the region above the $x$-axis bounded by the line $y=4x$ and the curve $y=x^3$? Find the area of the region above the $x$-axis bounded by the line $y=4x$ and the curve $y=x^3$
Attempt:
intersect when:-
$x^3 - 4x = 0$
$x ( x² - 4 ) = 0 $
$x = 0 , x = \pm 2$
Area is given by :-
$$2 ∫_0^2 4x - x^3\; dx = 2 [ 2x^2 - \frac13x^3 ]_0^2$$
$$2 [ 8 - 8/3 ] = 32/3\;\;\text{units²}$$
I want to understand this topic well so I'm solving different questions from textbooks under it. The answer in the textbook for this is 6.75 units². I can't figure where I'm wrong.
| The real problem is that $\displaystyle \int x^3 = \frac{x^4}{4}$, not $\displaystyle \frac{x^3}{3}$.
$$\int_0^2 4x-x^3$$
$$= \left[4 \cdot \frac{x^2}{2} - \frac{x^\color{red}{4}}{\color{red}{4}} \right]_0^2$$
$$= \left[ 2x^2 - \frac{x^4}{4} \right]_0^2$$
$$= \left(2\cdot 2^2 - \frac{2^4}{4} \right) - \left(2\cdot 0^2 - \frac{0^4}{4} \right)$$
$$ = (8 - 4) - (0)$$
$$ = 4$$
This agrees with the answer given by Wolfram Alpha, so your textbook is probably wrong. This shows that even textbooks make mistakes!
| {
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"timestamp": "2023-03-29T00:00:00",
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Let $A=\left\{a\in R:\text{the equation}(1+2i)x^3-2(3+i)x^2+(5-4i)x+2a^2=0\right\}$ has atleast one real root. Let $A=\left\{a\in R:\text{the equation}(1+2i)x^3-2(3+i)x^2+(5-4i)x+2a^2=0\right\}$ has atleast one real root.Find the value of $\sum_{a\in A}a^2$.
What should i do in this question to find the possible values of $a$.I have no clue how to start with.Please help me.Thanks.
| $2 a^2+x^3-6 x^2+i (2 x^3-2 x^2-4 x)+5 x = 0$
root of $2 x^3-2 x^2-4 x = 0$ are $-1, 0$ and $2$
$2 a^2+(-1)^3-6 (-1)^2+5 (-1) = 0$ imply $2a^2-12=0$ ie $a=\pm\sqrt{6}$
$2 a^2+(0)^3-6 (0)^2+5 (0) = 0$ imply $2a^2=0$ ie $a^2=0$
$2 a^2+(2)^3-6 (2)^2+5 (2) = 0$ imply $2a^2-6=0$ ie $a=\pm\sqrt{3}$
therefore $\sum a^2=18$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1473267",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Maximizing $3 x^2+2 \sqrt{2} x y$ with $x^4+y^4=1$ I want to have maxizing value of $3 x^2+2 \sqrt{2} x y$ when $x^4+y^4=1$, $x>0,y>0$.
How can I solve it.
| I have got solution with precaculus method. thank you everyone advices.
$3 x^2+2 \sqrt{2} x y$
$\leq$ $3 x^2+\left(x^2+2 y^2\right)$
$=$ $4 x^2+2 y^2$ (when equal is $x=\sqrt{2} y$ with arithmetic geometric mean inequality )
$=$ $2\cdot \left(2 x^2\right)+1\cdot \left(2
y^2\right)$
$\leq$ $\sqrt{\left(2^2+1^2\right) \left(\left(2
x^2\right)^2+\left(2
y^2\right)^2\right)}$
$=$ $\sqrt{\left(2^2+1^2\right) \left(4 x^4+4
y^4\right)}$ (when equal is $x=\sqrt{2} y$ with inequality of de Cauchy-Schwarz)
$=$ $\sqrt{5\cdot 4}$
$=$ $2 \sqrt{5}$
And I also can got the the solution with comment of @Théophile .
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Domain of functions that involve composition Suppose the domain of $f$ is $(-1,1)$. Define the function $\ell$ by
$$\ell(x)=f\left(\frac{x+1}{x-1}\right).$$ What is the domain of l
| As pointed in the comments, you want to solve: $$-1< \frac{x+1}{x-1} < 1$$for $x$. Rewrite $\frac{x+1}{x-1} = 1 + \frac{2}{x-1}$, so that: $$-1< \frac{x+1}{x-1} < 1 \implies -1 < 1 + \frac{2}{x-1} < 1 \implies -1 < \frac{1}{x-1} < 0.$$Can you solve that last one now?
| {
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"timestamp": "2023-03-29T00:00:00",
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Help showing $\lim\limits _{x\to -\infty \:}\left(\sqrt{4\cdot \:x^2-5\cdot \:x}+2\cdot \:x\right) = \frac{5}{4}$ I am stuck solving the following limit. I know the answer is 5/4, I just can't get it. This is the steps I have done so far.
$\lim _ \limits{x\to -\infty \:}\left(\sqrt{4\cdot \:x^2-5\cdot \:x}+2\cdot \:x\right)$
Multiply by Conjugate
$\lim _ \limits{x\to -\infty \:}\left(\sqrt{4\cdot \:x^2-5\cdot \:x}+2\cdot \:x\right)\cdot \frac{\left(\sqrt{4\cdot \:\:x^2-5\cdot \:\:x}-2\cdot \:\:x\right)}{\left(\sqrt{4\cdot \:\:x^2-5\cdot \:\:x}-2\cdot \:\:x\right)}$
Multiply Out
$\lim\limits_{x\to -\infty \:}\cdot \frac{\left(4\cdot \:\:\:x^2-5\cdot \:\:\:x-4\cdot \:\:x\right)}{\left(\sqrt{4\cdot \:\:x^2-5\cdot \:\:x}-2\cdot \:\:x\right)}$
Combine Like Terms
$\lim\limits_{x\to -\infty \:}\cdot \frac{\left(4\cdot \:\:\:x^2-9\cdot \:\:x\right)}{\left(\sqrt{4\cdot \:\:x^2-5\cdot \:\:x}-2\cdot \:\:x\right)}$
Factor out x
$\lim\limits_{x\to -\infty \:}\cdot \frac{x\left(4\cdot \:x-9\right)}{\left(\sqrt{x^2\left(4-\frac{5}{x}\right)}-2\cdot \:\:x\right)}$
Pull out x of sqrt and factor again
$\lim \limits_{x\to -\infty \:}\cdot \frac{x\left(4\cdot \:x-9\right)}{x\left(\sqrt{\left(4-\frac{5}{x}\right)}-2\right)}$
Now cancel x terms
$\lim \limits_{x\to -\infty \:}\frac{4\cdot \:\:x-9}{\sqrt{\left(4-\frac{5}{x}\right)}-2}$
Now I don't know what to do next.
If I plug in I get
$\frac{4\cdot \:\:\:-\infty \:-9}{\sqrt{\left(4-0\right)}-2}=\:\frac{-\infty \:}{0}$ Which doesn't equal $\frac{5}{4}$?
| Somehow, you missed a square.
$$
(\sqrt{4x^2-5x})^2-(2x)^2=4x^2-5x-4x^2=-5x
$$
which radically simplifies the ensuing limit calculation,
$$
\lim_{x\to-\infty}\frac{5|x|}{\sqrt{4x^2+5|x|}+2|x|}
=\lim_{x\to-\infty}\frac{5}{\sqrt{4+5/|x|}+2}=\frac54
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1475580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Find two numbers, $x,y$ whose sum is $35$ and $x^2y^5$ is maximum.
Find two numbers, $x,y$ whose sum is $35$ and $x^2y^5$ is maximum.
My answer:
$$x+y=35$$
$x^2y^5$ is maximum
$$y=35-x$$
$$\frac{d}{dx} x^2(35-x)^5$$
Which rule to apply here after? I reached:
$$(35-x)^4(-5x^2+(35-x)2x)=0$$
Either $(35-4x)^4 =0$ or $x^2-7=0$
| Another approach uses Lagrangian multipliers. Our Lagrangian is $L=2\ln x+5\ln y+\lambda (35-x-y)$ so $0=\partial_x L = \frac{2}{x}-\lambda,\,0=\partial_y L=\frac{5}{y}-\lambda$ and $x=10,\,y=25$. This maximises $x^2y^5$ because the matrix of second derivatives is diagonal, with negative eigenvalues $-\frac{2}{x^2},\,-\frac{5}{y^2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1475970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Show that $\lim_{n \longrightarrow \infty} \frac{3n^2+2n}{4+3n^2}=1$ I need show that $\forall \, \epsilon > 0$, exist $M \in \mathbb N$ such that, $\left|\frac{3n^2+2n}{4+3n^2}-1\right| < \epsilon$ if $n \geq M$.
Let's consider
$\left|\frac{3n^2+2n}{4+3n^2}-1\right|=\left|\frac{2n-4}{4+3n^2}\right|<\left|\frac{2n-4}{3n^2}\right|=\left|\frac{2}{3n}-\frac{4}{3n^2}\right|\leq \left|\frac{2}{3n^2}-\frac{4}{3n^2}\right|=\frac{2}{3n^2}<\epsilon \Leftrightarrow n>\sqrt{\frac{2}{3\epsilon}}$
Then, I must take $M = \left\lceil \sqrt{\frac{2}{3\epsilon}}\right\rceil+1 \in \mathbb{Z}$?
Thanks.
| $$\lim_{n \to \infty} \frac{3n^2 + 2n}{3n^2 + 2} = \lim_{n \to \infty} \frac{3 + 2/n}{3+2/n^2} = \frac{3}{3} = 1$$
$\epsilon$-proof:
$$ \left|\frac{3n^2 + 2n}{3n^2 + 2} - 1\right| = |\frac{2n - 2}{3n^2 + 2}| < \epsilon \Rightarrow |2n-2| < 3\epsilon n^2 + 2\epsilon$$
The absolute value can be removed since both sides are nonnegative when $n > 1$.
$$3\epsilon n^2 -2n + 2\epsilon + 2 > 0$$
This is a parabola with the opening facing up. Using the quadratic formula,
$$n > \frac{2 + \sqrt{2^2 - 4 \cdot 3\epsilon \cdot (2\epsilon + 2)}}{6\epsilon} = \frac{1 + \sqrt{1 - 6\epsilon - 6\epsilon^2}}{3\epsilon} = M$$
$\pm$ is changed to $+$ since the other root is negative.
| {
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"timestamp": "2023-03-29T00:00:00",
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Computing $\int \frac{4\tan(x)+5}{\sin^2(x)+2\cos^2(x)+3\sin(x)\cos(x)} $ $$\int \frac{4\tan(x)+5}{\sin^2(x)+2\cos^2(x)+3\sin(x)\cos(x)} $$
This question was asked today in my maths exam,
It was one of those two questions which I couldn't answer, How do you go about answering it ?
| $$\begin{align}\int \frac{4\tan(x)+5}{\sin^2(x)+2\cos^2(x)+3\sin(x)\cos(x)} dx
&=
\int \frac{4\tan(x)+5}{\cos^2(x)(\tan (x)+1) (\tan (x)+2)}dx\\
&=
\int \frac{4u+5}{(u+1) (u+2)}du
\end{align}$$
The last part is by using $u=\tan x$, $du = \frac{dx}{\cos^2 x}$.
I hope you know to continue from here.
Note:
$$
\sin^2(x)+2\cos^2(x)+3\sin(x)\cos(x) =\cos^2(x)( \tan^2(x)+2+3\tan(x))
$$
denote $\tan x = u$ to get
$\tan^2(x)+2+3\tan(x) = u^2+2+3u = (u+1)(u+2)$
| {
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"timestamp": "2023-03-29T00:00:00",
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Convergence of $\sum_{n=1}^{\infty} \log\left(\frac{(2n)^2}{(2n+1)(2n-1)}\right)$ I have to show that the series
$\sum_{n=1}^{\infty} \log\left(\frac{(2n)^2}{(2n+1)(2n-1)}\right)$
converges.
I have tried Ratio Test and Cauchy Condensation Test but it didn't work for me. I tried using Comparison Test but I couldn't make an appropriate inequality for it. Could you please give me some hints. Any help will be appreciated.
| Since $\log (1 + x) < x$ for non-negative $x$, we have
$$\begin {eqnarray}
0 < \sum_{n = 1}^{\infty} \log \left (\frac {(2n)^2} {(2n - 1) (2n + 1)} \right) & = & \sum_{n = 1}^{\infty} \log \left (1 + \frac {1} {(2n - 1) (2n + 1)} \right) \nonumber \\ & < & \sum_{n = 1}^{\infty} \left ( \frac {1} {(2n - 1) (2n + 1)} \right) \nonumber \\ & = & \frac {1} {2} \left (\sum_{n = 1}^{\infty} \frac {1} {2n - 1} - \sum_{n = 1}^{\infty} \frac {1} {2n + 1} \right) \nonumber \\ & = & \frac {1} {2} \left (\sum_{n = 0}^{\infty} \frac {1} {2n + 1} - \sum_{n = 1}^{\infty} \frac {1} {2n + 1} \right) \nonumber \\ & = & \frac {1} {2} \cdot 1 = \frac {1} {2}.
\end {eqnarray}$$
So, the series converges.
| {
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How to prove this $\sqrt{x^2+3}+\sqrt{y^2+3}+\sqrt{xy+3}\ge 6$ Let $x,y>0$, and $x+y=2$, show that
$$\sqrt{x^2+3}+\sqrt{y^2+3}+\sqrt{xy+3}\ge 6$$
I tried using Minkowski inequality
$$\sqrt{x^2+3}+\sqrt{y^2+3}\ge\sqrt{(x+y)^2+(\sqrt{3}+\sqrt{3})^2}=\sqrt{4+12}=4$$
But
$$\sqrt{xy+3}\le\sqrt{\dfrac{(x+y)^2}{4}+3}=2$$
so i'm not sure how it would work with this problem,Thanks
| To prove that
$\sqrt{x^2+3}+\sqrt{y^2+3}+\sqrt{xy+3}\ge 6$,
We use Jensen's inequality.
$1*\sqrt{x^2+3}+1*\sqrt{y^2+3}+1*\sqrt{xy+3}\ge (1+1+1) \sqrt{\frac{x^2+y^2+xy+9}
{1+1+1}}$
Then, since $x+y=2$, we have:
$\sqrt{x^2+3}+\sqrt{y^2+3}+\sqrt{xy+3} \ge 3\sqrt{\frac{x^2+(2-x)^2+x(2-x)+9}{3}}$
$\sqrt{x^2+3}+\sqrt{y^2+3}+\sqrt{xy+3} \ge 3\sqrt{\frac{x^2-2x+13}{3}} = 3\sqrt{\frac{(x-1)^2+12}{3}} \ge 6$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1480883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
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} |
Find the sum from the system of equations
If $x,y, z$ satisfy: $$x + y = z^2 + 1, y + z = x^2 + 1, x + z = y^2 + 1 $$ Find the value of $2x +3y + 4z$.
This gives us (by getting $x + y + z$ that)
$z^2 + z + 1 = x^2 + x + 1 = y^2 + y + 1 \implies z^2 + z = x^2 + x = y^2 + y$.
Using the first and last, I also got:
$2x + y + z = z^2 + y^2 + 2$
But I cannot get much farther!
EDIT:
$x -z = z^2 - x^2 = (z-x)(z+x) \implies z + x = -1$.
$y - x = x^2 = y^2 = (x-y)(x+y) \implies x + y = -1$
Thus, $z - y = 0$ and $z = y$.
$2x + 7y$ is to be found then, and $x = -1 -y $ so: $2(-1 - y) - 7y = -2 - 9y$ is to be found.
Any ideas here?
| You can add up the three equations to get
$2x+2y+2z = x^2+y^2+z^2 +3
= (x^2+1)+ (y^2+1)+ (z^2+1)\geq 2x+2y+2z$
Hence, we know that $x=y=z=1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1482063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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proving montonity and convergence of sequence en = (1 + 1/n)^n Prove the following. What would be the summation formula be for the first part?
| From the Binomial Theorem, we have
$$\begin{align}
e_n&=\left(1+\frac1n\right)^n\\\\
&=\sum_{k=0}^n\binom{n}{k}\left(\frac1n\right)^k\\\\
&=1+n\left(\frac1n\right)+\frac{n(n-1)}{2!}\left(\frac1n\right)^2+\cdots+\frac{(n)(n-1)(n-2)\cdots(n-(n-1))}{n!}\left(\frac1n\right)^n\\\\
&=1+1+\frac1{2!}\left(1-\frac1n\right)+\frac1{3!}\left(1-\frac1n\right)\left(1-\frac2n\right)\\\\
&+\cdots+\frac1{n!}\left(1-\frac1n\right)\left(1-\frac2n\right)\cdots\left(1-\frac{n-1}n\right) \tag 1
\end{align}$$
as was to be shown!
To show that $e_n$ is increasing we will show that $\frac{e_{n+1}}{e_n}\ge 1$. To that end, we write
$$\begin{align}
\frac{e_{n+1}}{e_n}&=\frac{\left(1+\frac{1}{n+1}\right)^{n+1}}{\left(1+\frac{1}{n}\right)^n}\\\\
&=\left(\frac{n(n+2)}{(n+1)^2}\right)^{n+1}\left(1+\frac1n\right)\\\\
&=\left(1-\frac{1}{(n+1)^2}\right)^{n+1}\left(1+\frac1n\right) \tag 2\\\\
&\ge\left(1-\frac{1}{n+1}\right)\left(1+\frac1n\right) \tag 3\\\\
&=1
\end{align}$$
where in going from $(2)$ to $(3)$, we used Bernoulli's Inequality. Thus, $e_n$ in monotonically increasing; and we are done!
From $(1)$ is trivial to see that $e_n\ge 2$ for $n\ge 1$.
Noting that all of the parenthetical terms are less than $1$, and $n!>2^{n-1}$ for $n\ge2$, then for $n\ge 2$, we have
$$e_n\le 1+1+\sum_{k=1}^\infty \frac{1}{2^k}=3$$
Thus, we conclude that $e_n$ is a monotonically increasing sequence that is bounded above by $3$. Therefore, $e_n$ converges to some real number betweem $2$ and $3$. And we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1482253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Logarithmic inequality and properties of logarithms So the inequality:
$$2 \cdot \log_{\sqrt3}{(1-x)} - \log_\sqrt3 {(3-x)} \lt 2$$
Can be written as:
$$\log_{\sqrt3}{(1-x)^2} - \log_\sqrt3 {(3-x)} \lt 2$$
?????????????
I have tried both on Wolfram|Alpha and both gave different intervals of x, although the logarithm property of exponent is: $\log_a{b}^n = n \cdot \log_a{b}$
The interval of solutions for the first inequality is:
$$\frac{-1-\sqrt{33}}{2} \lt x \lt 1$$
While for the second one:
$$\frac{-1-\sqrt{33}}{2} \lt x \lt 1 \quad \lor \quad 1 \lt x \lt \frac{\sqrt{33}-1}{2}$$
I guess the reason is the different field of existence "I am not sure what's it called in English, it's basically $\log_a{f(x)} \qquad f(x) \gt 0$ to make the logarithm valid". Or in other words...
First inequality:
$$\begin{cases}
\frac{-1-\sqrt{33}}{2} \lt x \lt \frac{\sqrt{33}-1}{2} \lor x \gt 3 & \mbox{Solution of inequality between the arguments of the logarithms}\\ x \lt 1 & \mbox{Condition of the first argument} \\ x \lt 3 & \mbox{Condition of the second argument}\end{cases} $$
Second one:
$$\begin{cases}
\frac{-1-\sqrt{33}}{2} \lt x \lt \frac{\sqrt{33}-1}{2} \lor x \gt 3 & \mbox{Solution of inequality between the arguments of the logarithms}
\\ (1-x)^2 \gt 0 & \mbox{Condition of the first argument}
\\ x \lt 3 & \mbox{Condition of the second argument}
\end{cases}$$
So it will be:
$$
\begin{cases}
\frac{-1-\sqrt{33}}{2} \lt x \lt \frac{\sqrt{33}-1}{2} \lor x \gt 3
\\ x \neq 1
\\ x \lt 3
\end{cases}
$$
So the question is why that difference happens though I only used a logarithmic property, and how to decide the values of {$x$} to be sure that the logarithm is valid, I mean why can't I choose $(1-x)^2 \gt 0$ in the first one too since they're basically the same?
| Note that throughout, $x\neq 1$ and $x\neq 3$ applies.
$$\begin{align}
2\log_\sqrt{3}(1-x)-\log_\sqrt{3}(3-x)&<2\\
\log_\sqrt{3}\frac{(1-x)^2}{3-x}&<\log_\sqrt{3}3\\
\frac {(1-x)^2}{3-x}&<3\\
\frac {(1-x)^2}{3-x}-3&<0\\
\frac {x^2+x-8}{3-x}&<0\\
\frac {x^2+x-8}{x-3}&>0\\
\frac {(x-\alpha)(x-\beta)}{x-3}&>0\\
\text{where $\alpha=\frac {-1-\sqrt{33}}2=-3.372, \beta=\frac{-1+\sqrt{33}}2=2.372$}&\\
\begin{cases}-3.372<x<2.372\;\; (x\neq 1) \\x>3\end{cases}\quad\blacksquare\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1483559",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove by induction: $\sum\limits_{k=1}^{n}sin(kx)=\frac{sin(\frac{n+1}{2}x)sin\frac{nx}{2}}{sin\frac{x}{2}}$ $\sum\limits_{k=1}^{n}sin(kx)=\frac{sin(\frac{n+1}{2}x)sin\frac{nx}{2}}{sin\frac{x}{2}}$
Base case: For $n=1$
$sinx=\frac{sinx\cdot sin\frac{x}{2}}{sin\frac{x}{2}}=sinx$
Induction hypothesis: For $n=m$
$\sum\limits_{k=1}^{m}sin(kx)=\frac{sin(\frac{m+1}{2}x)sin\frac{mx}{2}}{sin\frac{x}{2}}$
Induction step: $n=m+1$
$\sum\limits_{k=1}^{m+1}sin(kx)=\frac{sin(\frac{m+2}{2}x)sin\frac{(m+1)x}{2}}{sin\frac{x}{2}}$
Prove: $\frac{sin(\frac{m+1}{2}x)sin\frac{mx}{2}}{sin\frac{x}{2}}+sin(m+1)x=\frac{sin(\frac{m+2}{2}x)sin\frac{(m+1)x}{2}}{sin\frac{x}{2}}$
Left side: $\frac{sin(\frac{m+1}{2}x)sin\frac{mx}{2}+sin\frac{x}{2}sin(m+1)x}
{sin\frac{x}{2}}$
How to prove this equality? I used $sin(u)sin(v)$ identity but that didn't help.
| Notice, in the third step of induction, you should follow the right procedure
substituting $n=m+1$ in the equality, we get
$$\sum_{k=1}^{m+1}\sin(kx)=\frac{\sin\left(\frac{(m+2)x}{2}\right)\sin\left(\frac{(m+1)x}{2}\right)}{\sin\left(\frac{x}{2}\right)}$$
$$\sum_{k=1}^{m}\sin(kx)+\sin(m+1)x=\frac{\sin\left(\frac{(m+2)x}{2}\right)\sin\left(\frac{(m+1)x}{2}\right)}{\sin\left(\frac{x}{2}\right)}$$
$$\sum_{k=1}^{m}\sin(kx)=\frac{\sin\left(\frac{(m+2)x}{2}\right)\sin\left(\frac{(m+1)x}{2}\right)}{\sin\left(\frac{x}{2}\right)}-\sin(m+1)x$$
$$\sum_{k=1}^{m}\sin(kx)=\frac{\sin\left(\frac{(m+2)x}{2}\right)\sin\left(\frac{(m+1)x}{2}\right)-\sin(m+1)x\sin\left(\frac{x}{2}\right)}{\sin\left(\frac{x}{2}\right)}$$
Applying $\color{red}{\sin 2A=2\sin A\cos A}$
$$=\frac{\sin\left(\frac{(m+2)x}{2}\right)\sin\left(\frac{(m+1)x}{2}\right)-2\sin\left(\frac{(m+1)x}{2}\right)\cos\left(\frac{(m+1)x}{2}\right)\sin\left(\frac{x}{2}\right)}{\sin\left(\frac{x}{2}\right)}$$
$$=\frac{\sin\left(\frac{(m+1)x}{2}\right)}{\sin\left(\frac{x}{2}\right)}\left(\sin\left(\frac{mx}{2}+x\right)-2\cos\left(\frac{mx}{2}+\frac{x}{2}\right)\sin\left(\frac{x}{2}\right)\right)$$
Applying $\color{red}{2\cos A\sin B=\sin(A+B)-\sin(A-B)}$
$$=\frac{\sin\left(\frac{(m+1)x}{2}\right)}{\sin\left(\frac{x}{2}\right)}\left(\sin\left(\frac{mx}{2}+x\right)-\left(\sin\left(\frac{mx}{2}+\frac{x}{2}+\frac{x}{2}\right)-\sin\left(\frac{mx}{2}+\frac{x}{2}-\frac{x}{2}\right)\right)\right)$$
$$=\frac{\sin\left(\frac{(m+1)x}{2}\right)}{\sin\left(\frac{x}{2}\right)}\left(\sin\left(\frac{mx}{2}+x\right)-\sin\left(\frac{mx}{2}+x\right)+\sin\left(\frac{mx}{2}\right)\right)$$
$$\sum_{k=1}^{m}\sin(kx)=\frac{\sin\left(\frac{(m+1)x}{2}\right)\sin\left(\frac{mx}{2}\right)}{\sin\left(\frac{x}{2}\right)}$$
Which is true by the hypothesis.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1484286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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question related with half angle from a given equation If $\sec x + \tan x = 3$ , then what is the value of $\tan \frac x2?$
i squared both sides and then converted the $\tan x $ into $\sec x$ and then $\sec x $ into $\cos x$ which gave me value $\sqrt\frac15$. Kinda confused after this step.. help is appreciated
| HINT:
$$\sec(x)+\tan(x)=3\Longleftrightarrow$$
$$-3+\sec(x)+\tan(x)=0\Longleftrightarrow$$
Substitute $y=\tan\left(\frac{x}{2}\right)$. Than $\sin(x)=\frac{2y}{y^2+1}$ and $\cos(x)=\frac{1-y^2}{y^2+1}$:
$$\frac{2}{y-1}-\frac{4y}{y-1}=0\Longleftrightarrow$$
$$-\frac{2(2y-1)}{y-1}=0\Longleftrightarrow$$
$$\frac{2y-1}{y-1}=0\Longleftrightarrow$$
$$2y=1\Longleftrightarrow$$
$$y=\frac{1}{2}\Longleftrightarrow$$
$$\tan\left(\frac{x}{2}\right)=\frac{1}{2}\Longleftrightarrow$$
$$\frac{x}{2}=\tan^{-1}\left(\frac{1}{2}\right)+\pi n\Longleftrightarrow$$
$$x=2\tan^{-1}\left(\frac{1}{2}\right)+2\pi n$$
With $n\in\mathbb{Z}$
So it if you take $n=0$:
$$x=2\tan^{-1}\left(\frac{1}{2}\right)+2\pi n \Longrightarrow$$
$$x=2\tan^{-1}\left(\frac{1}{2}\right)+2\pi \cdot 0 =2\tan^{-1}\left(\frac{1}{2}\right) $$
Than $\tan\left(\frac{x}{2}\right)$ is:
$$\tan\left(\frac{x}{2}\right) \Longrightarrow$$
$$\tan\left(\frac{2\tan^{-1}\left(\frac{1}{2}\right)}{2}\right)=\tan\left(\tan^{-1}\left(\frac{1}{2}\right)\right)=\frac{1}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1484609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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$2^\text{nd}$ Derivative of normal distribution, evaluated at one standard deviation What is the $2^{nd}$ derivative of the normal distribution at one standard deviation?
The normal distribution is given by $N(x,\mu ,\sigma)=\frac{1}{\sigma\sqrt{2\pi }}e^{-\frac{(x-\mu)^2}{2\sigma^2}}$. To make this problem easier, lets say I have a standard normal distribution($\mu =0,\sigma =1$). So $N\left(x,0,1\right)=\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}$.
$$\frac{d}{dx}N(x,0,1)=\frac{d}{dx}\left(\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}\right)$$
$$\frac{d}{dx}N\left(x,0,1\right)=\frac{1}{\:\sqrt{2\pi}}\frac{d}{dx}\left(e^{-\frac{x^2}{2}}\right)$$
$$\frac{d}{dx}N\left(x,0,1\right)=\frac{1}{\:\sqrt{2\pi}}e^{-\frac{\left(x\:\right)^2}{2}\cdot \frac{d}{dx}\left(-\frac{x^2}{2}\right)}$$
$$\frac{d}{dx}N\left(x,0,1\right)=\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}\cdot -x}$$
So, $$\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}\cdot\:-x}$$ is the first derivative. To get the second, I took the derivative of the first.
$$\frac{d}{dx}\left(\frac{d}{dx}N\left(x,0,1\right)\right)=\frac{d}{dx}\left(\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}\cdot \:-x}\right)$$
$$\frac{d}{dx}\left(\frac{d}{dx}N\left(x,0,1\right)\right)=\frac{1}{\:\sqrt{2\pi }}\frac{d}{dx}\left(e^{-\frac{x^2}{2}\cdot\:-x}\right)$$
$$\frac{d}{dx}\left(\frac{d}{dx}N(x,0,1)\right)=\frac{1}{\:\sqrt{2\pi}}e^{-\frac{x^2}{2}\cdot\:-x}\cdot \frac{d}{dx\:}\left(-\frac{x^2}{2}\cdot \:-x\right)$$
$$\frac{d}{dx}\left(\frac{d}{dx}N\left(x,0,1\right)\right)=\frac{1}{\sqrt{2\pi}}\left(e^{-\frac{(x^2}{2}\cdot \:-x}\right)\cdot \left(3\frac{x^2}{2}\right)$$
So now I evaluate $$\frac{1}{\sqrt{2\pi}}\left(e^{-\frac{x^2}{2}\cdot -x}\right)\cdot \left(3\cdot \frac{x^2}{2}\right),$$ at the standard deviation which I set to $\sigma =1$
So, $$\frac{1}{\sqrt{2\cdot :\pi}}\left(e^{-\frac{1^2}{2}\cdot -1}\right)\cdot \left(3\cdot\frac{1^2}{2}\right)$$
$$\frac{1}{\sqrt{2\pi}}\left(e^{-\frac{(1)^2}{2}\cdot\,-1}\right)\cdot \left(3\cdot \frac{1^2}{2}\right)=\frac{3\sqrt{e}}{2\sqrt{2\cdot\pi}}$$
But my teacher says the answer is suppose to be $0$? What am I doing wrong? Side Note: I am new to Calculus. So an elaborate explanation will be appreciated.
| Your way of applying the chain rule is wrong. Here's the right way:
$$
\frac d {dx} e^{-x^2/2} = e^{-x^2/2} \cdot \frac d {dx} \left( \frac{-x^2} 2 \right) = e^{-x^2/2} \cdot(-x).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1485036",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Show trigonometric identity $\frac{1-\tan\frac{\theta}{2}}{1+\tan\frac{\theta}{2}} = \sec\theta - \tan\theta$ I need to show that $\frac{1-\tan\frac{\theta}{2}}{1+\tan\frac{\theta}{2}} = \sec\theta - \tan\theta$
So far, I have changed the fraction to (where $t=\tan\frac{\theta}{2}$), $\frac{1-2t+t^2}{1-t^2}$(*)
I know that $\tan\theta = \frac{2\tan\frac{\theta}{2}}{1-\tan^2\frac{\theta}{2}}$ and that I should be able to use this to complete the proof, but I am unsure how to split the fraction (*) I have arrived at in order to utilise this?
| Notice, $$\frac{1-\tan\frac{\theta}{2}}{1+\tan\frac{\theta}{2}}=\frac{\left(1-\tan\frac{\theta}{2}\right)\left(1+\tan\frac{\theta}{2}\right)}{\left(1+\tan\frac{\theta}{2}\right)^2}=\frac{1-\tan^2\frac{\theta}{2}}{1+\tan^2\frac{\theta}{2}+2\tan\frac{\theta}{2}}$$
$$=\frac{1}{\frac{1+\tan^2\frac{\theta}{2}}{1-\tan^2\frac{\theta}{2}}+\frac{2\tan\frac{\theta}{2}}{1-\tan^2\frac{\theta}{2}}}=\frac{1}{\sec \theta+\tan \theta}=\frac{\sec\theta-\tan \theta}{\sec^2 \theta-\tan^2\theta}=\sec\theta-\tan \theta$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1485614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to prove that the following system of equations has only one solution? $
\begin{cases}
(x - 1)^2 + (y + 1)^2 = 25 \\
(x + 5)^2 + (y + 9)^2 = 25 \\
y = -\frac{3}{4}x - \frac{13}{2}
\end{cases}
$
I have to solve this system of equations. After substituting $y = -\frac{3}{4}x - \frac{13}{2}$ into $(x - 1)^2 + (y + 1)^2 = 25$, I got that $(x + 2)^2 = 0$, so $x = -2$, so $y = -\frac{3}{4} \cdot (-2) - \frac{13}{2} = -5$.
How to prove that this is the only solution?
| I assume that you have checked that $(x,y) = (-2,-5)$ is indeed a solution.
Why is it the only one?
It is, because you have shown that your three equations must imply that $(x+2)^2=0$, which in turn implies $x=-2$. In other words, whatever solutions $(x,y)$ the three original equations have, you have shown that they must all have the property that $x=-2$.
But given that $x=-2$, the last of the three equations show that we must have $y=-5$. In conclusion, if there is a solution at all to your three equations, then it must be $(x,y)=(-2,-5)$ (and you do need to check that this is a solution, because you can easily end up in the case where there are no solutions at all).
| {
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"timestamp": "2023-03-29T00:00:00",
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Why test of divisible by $12$ works with $3$ and $4$ but not with $2$ and $6 ?$ Test of divisible by $4 ,$ last two digit must be divisible by $4 ,$ since $100$ is always divisible by $4$ remaining two digit $,$ we need to check $.$
Test of divisible by $3 ,$ sum digits must be divisible by $3 .$ But I don't know why $?$
Test of divisible by $2 ,$ last digit must be divisible by $2 ,$ since $10$ is always divisible by $2 .$
Test of divisible by $ 6 ,$ the number should be divisible by $2$ and $3 .$
Test of divisible by $12 ,$ the number should be divisible by $4$ and $3 ,$ somewhere is given as we can say if number is divisible by $2$ and $6$ then it also must be divisible by $12 ,$ for counter example $,$ if number is $18 .$
I have two questions $:$
*
*Why test of divisible by $3$ works $?$ Any proof $.$
*Why test of divisible by $12$ works with divisible by $3$ and $4 ,$ but not with divisible by $2$ and $6 ?$ Any proof $.$
| The reason the divisibility by $3$ test works is that $10$ and $1$ have the same remainder when divided by $3$. Say your number is (as digits) $x = a_n a_{n-1} \ldots a_1 a_0$. Then you could write it as $x = a_n \cdot 10^n + a_{n-1} 10^{n-1} + \cdots + a_1 10 + a_0$. If you mod out by $3$:
$$
\begin{align*}
x \bmod 3 &= (a_n \cdot 10^n + a_{n-1} 10^{n-1} + \cdots + a_1 10 + a_0) \bmod 3 \\
&= (a_n \cdot 10^n) \bmod 3 + (a_{n-1} 10^{n-1}) \bmod 3 + \cdots + (a_1 10) \bmod 3 + a_0 \bmod 3) \\
&= (a_n \bmod 3) \cdot (10^n \bmod 3) + (a_{n-1} \bmod 3) (10^{n-1} \bmod 3) + \cdots + (a_1 \bmod 3) \cdot (10 \bmod 3) + (a_0 \bmod 3) \\
&= (a_n \bmod 3) \cdot 1 + (a_{n-1} \bmod 3) \cdot 1 + \cdots + (a_1 \bmod 3) \cdot 1 + (a_0 \bmod 3) \\
&= (a_n \bmod 3) + (a_{n-1} \bmod 3) + \cdots + (a_1 \bmod 3) + (a_0 \bmod 3) \\
&= (a_n + a_{n-1} + \cdots + a_1 + a_0) \bmod 3
\end{align*}
$$
So $x$ and the sum of its digits have the same remainder when divided by $3$.
As for your second question: if you know $x$ is divisible by $6$, you already know it's divisible by $2$. Doing the test for $2$ doesn't give you any new information. This is because $6$ and $2$ aren't relatively prime (but $3$ and $4$ are). In general, if you know $x$ is divisible by $a$ and $b$, it's divisible by the LCM of $a$ and $b$.
| {
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"timestamp": "2023-03-29T00:00:00",
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A serie about $\sum_{n = 1}^\infty {\arctan \frac{{10n}}{{\left( {3{n^2} + 2} \right)\left( {9{n^2} - 1} \right)}}}$ How to prove $$\sum\limits_{n = 1}^\infty {\arctan \frac{{10n}}{{\left( {3{n^2} + 2} \right)\left( {9{n^2} - 1} \right)}}} = \ln 3 - \frac{\pi }{4}.$$
Add: Maybe we can follow this!
| \begin{align*}S&=\sum\limits_{n=1}^{\infty} \arctan \frac{10n}{(3n^2+2)(9n^2-1)} \\&= \sum\limits_{n=1}^{\infty} \arg \left(1+\frac{10in}{(3n^2+2)(9n^2-1)}\right)\\&= \arg \prod\limits_{n=1}^{\infty}\left(1+\frac{10in}{(3n^2+2)(9n^2-1)}\right)\\&= \arg \prod\limits_{n=1}^{\infty}\left(\frac{(3n^2+2)(9n^2-1)+10in}{27n^4\left(1+\frac{2}{3n^2}\right)\left(1-\frac{1}{9n^2}\right)}\right)\end{align*}
The products in the denominator $\displaystyle \prod\limits_{n=1}^{\infty}\left(1+\frac{2}{3n^2}\right)$ and $\displaystyle \prod\limits_{n=1}^{\infty}\left(1-\frac{1}{9n^2}\right)$ can be ignored as they are convergent and real.
So, \begin{align*}S&= \arg \prod\limits_{n=1}^{\infty}\left(\frac{(3n^2+2)(9n^2-1)+10in}{27n^4}\right)\end{align*}
Using the factorization of the numerator,
$$(3n^2+2)(9n^2-1)+10in = (n-i)(3n+i)(3n+i+1)(3n+i-1)$$ we get,
$$S = \arg \prod\limits_{n=1}^{\infty}\frac{\left(1+\frac{i}{3n}\right)\left(1+\frac{i+1}{3n}\right)\left(1+\frac{i-1}{3n}\right)}{\left(1+\frac{i}{n}\right)}$$
Using, $\displaystyle \frac{1}{\Gamma(z)} = ze^{\gamma z}\prod\limits_{n=1}^{\infty}\left(1+\frac{z}{n}\right)e^{-z/n}$ at $\displaystyle z = i,\frac{i}{3},\frac{i+1}{3},\frac{i-1}{3}$ we may rewrite it as,
$$S = \arg \frac{-\Gamma(i)}{\Gamma\left(\frac{i}{3}\right)\Gamma\left(\frac{i+1}{3}\right)\Gamma\left(\frac{i-1}{3}\right)}$$
On the other hand we have Gauss-Legendre Triplication Formula:
$\displaystyle \Gamma(3z) = \frac{1}{2\pi}3^{3z - \frac{1}{2}}\Gamma\left(z\right)\Gamma\left(z+\frac{1}{3}\right)\Gamma\left(z+\frac{2}{3}\right)$,
which at $z = \dfrac{i-1}{3}$ gives:
$$\Gamma\left(\frac{i-1}{3}\right)\Gamma\left(\frac{i}{3}\right)\Gamma\left(\frac{i+1}{3}\right) = 2\pi 3^{-i+\frac{3}{2}}\Gamma(i-1)$$
Hence, $$S = \arg \frac{-3^{i}\Gamma(i)}{\Gamma(i-1)} = \arg (3^{i}(1-i)) = \log 3 - \frac{\pi}{4}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Finding equation of chord of hyperbola. Equation of chord of hyperbola joining points $(a\sec\phi,b\tan\phi)$ and $(a\sec\phi_1,b\tan\phi_1) $ $$y-b\tan\phi=\frac{b\tan\phi-b\tan\phi_1}{a\sec\phi-asec \phi_1}(x-a\sec\phi) $$ This reduces to $$\frac{y}{b}\sin\Big(\frac{\phi+\phi_1}{2}\Big)-\frac{x}{a}\cos\Big(\frac{\phi-\phi_1}{2}\Big)=\tan\phi \sin\Big(\frac{\phi+\phi_1}{2}\Big)-\sec\phi \cos\Big(\frac{\phi-\phi_1}{2}\Big) $$
Now I want to reduce this in the form$$-\frac{y}{b}\sin\Big(\frac{\phi+\phi_1}{2}\Big)+\frac{x}{a}\cos\Big(\frac{\phi-\phi_1}{2}\Big)=\cos\Big(\frac{\phi+\phi_1}{2}\Big) $$
How to get this form ? Most of the books use this one but they dont give the proof which I seek.
| Another method which does not require ingenious use of trigonometric identities goes as follows
Equation of tangent to hyperbola at point $(asec \ A,btan \ A)$ is $$\frac{x}{a}sec \ A-\frac{y}{b}tan\ A=1 $$
Equation of tangent to hyperbola at point $(asec \ B,btan \ B)$ is $$\frac{x}{a}sec \ B-\frac{y}{b}tan\ B=1 $$
The intersection of these two tangents is the point $$\Bigg(a\frac{cos\frac{A-B}{2}}{cos\frac{A+B}{2}},b\frac{sin\frac{A+B}{2}}{cos\frac{A+B}{2}}\Bigg) $$
The equation of chord of contact from a point on a conic is $T=0$. Hence equation of chord is $$\frac{x}{a^2}a\frac{cos\frac{A-B}{2}}{cos\frac{A+B}{2}} -\frac{y}{b^2}b\frac{sin\frac{A+B}{2}}{cos\frac{A+B}{2}}=1 $$ Which on simplification gives $$\frac{x}{a}{cos\frac{A-B}{2}}-\frac{y}{b}sin\frac{A+B}{2}=cos\frac{A-B}{2} $$
| {
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Of sum of cosines and the $7$th roots of unity In my solution here, it was shown that
$$\omega+\omega^2+\omega^4=-\frac 12\pm\frac{\sqrt7}2\qquad\qquad (\omega=e^{i2\pi/7})$$
from which we know that
$$\sin \frac{2\pi}7+\sin \frac{4\pi}7-\sin \frac{6\pi}7=\Im (\omega+\omega^2+\omega^4)=\frac{\sqrt7}2$$
This can also be verified easily by computation.
By the same token it would appear that
$$\cos \frac{2\pi}7+\cos \frac{4\pi}7-\cos \frac{6\pi}7=\Re (\omega+\omega^2+\omega^4)=\frac 12$$
However a quick computational check shows that the result is $1.3019...$ and not $\frac 12$.
Why is this so?
| You're starting from the wrong assumption that $-\omega^3=\omega^4$, which is obviously wrong, because it would imply $\omega=-1$.
The true fact is that
$$
-\sin\frac{6\pi}{7}=\sin\frac{8\pi}{7}
$$
but
$$
\cos\frac{6\pi}{7}=\cos\frac{8\pi}{7}
$$
because
$$
\frac{6\pi}{7}+\frac{8\pi}{7}=2\pi
$$
and
$$
\sin(2\pi-\alpha)=-\sin\alpha,\qquad
\cos(2\pi-\alpha)=\cos\alpha
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1491566",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How can I convert this tricky complex number into a real number: $\int_0^{\infty}\frac{x^α}{x^3+1}dx$? The problem statement is:
$$\int_0^{\infty}\frac{x^α}{x^3+1}dx$$
for α in the range −1<α<2.
$$\huge \frac{2\pi i}{1-e^{\frac{i2\pi (\alpha+1)}{3}}} \frac {e^{\frac{i \pi \alpha}{3}}} { 3e^{\frac{2\pi i}{3}}}$$
$\alpha$ is some constant between -1 and 2.
Thanks,
| HINT:
Assuming $x\in\mathbb{R}$:
$$\frac{2\pi i}{1-e^{\frac{i2\pi x}{3}}} \frac {e^{\frac{i \pi x}{3}}} { 3e^{\frac{2\pi i}{3}}}=$$
$$\frac{\left(2\pi i\right)\cdot\left(e^{\frac{x\pi i}{3}}\right)}{\left(1-e^{\frac{i2\pi x}{3}}\right)\cdot\left(3e^{\frac{2\pi i}{3}}\right)}=$$
$$\frac{2\pi i e^{\frac{x\pi i}{3}}}{\left(1-e^{\frac{i2\pi x}{3}}\right)\cdot\left(3e^{\frac{2\pi i}{3}}\right)}=$$
$$\frac{2\pi i e^{\frac{x\pi i}{3}}}{\left(1-e^{\frac{i2\pi x}{3}}\right)\cdot\left(3e^{\frac{2\pi i}{3}}\right)}=$$
$$\frac{2\pi i\left(\cos\left(\frac{\pi x}{3}\right)+\sin\left(\frac{\pi x}{3}\right)i\right)}{\left(1-e^{\frac{i2\pi x}{3}}\right)\cdot\left(3e^{\frac{2\pi i}{3}}\right)}=$$
$$\frac{-2\pi\sin\left(\frac{\pi x}{3}\right)+2i\pi\cos\left(\frac{\pi x}{3}\right)}{\left(1-e^{\frac{i2\pi x}{3}}\right)\cdot\left(3e^{\frac{2\pi i}{3}}\right)}=$$
$$\frac{-2\pi\sin\left(\frac{\pi x}{3}\right)+2i\pi\cos\left(\frac{\pi x}{3}\right)}{3(i+\sqrt{3})e^{\frac{i\pi x}{3}}\sin\left(\frac{\pi x}{3}\right)}$$
| {
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How to Calculate $x^6+x^3y^3+y^6$ Given that $x,y$ real numbers such that :
$x^2+xy+y^2=4$
And
$x^4+x^2y^2+y^4=8$
How can one calculate :
$x^6+x^3y^3+y^6$
Can someone give me hint .
| Denote $a=x/y$, $b=xy$.
Then
$$a+1+\frac{1}{a}=\frac{4}{b},$$
$$a^2+1+\frac{1}{a^2}=\frac{8}{b^2},$$
$$a^3+1+\frac{1}{a^3}=?$$
Then from the $1$st equation:
$$
\left(a+\frac{1}{a}\right)^2 = \left(\frac{4}{b}-1\right)^2,
$$
combining it with $2$nd equation, $\Rightarrow$ $b=1$.
Then, $a+\frac{1}{a}=3$, $a^2+\frac{1}{a^2}=7$; therefore $a^3+\frac{1}{a^3}=\left(a+\frac{1}{a}\right)\left(a^2+\frac{1}{a^2}\right) - \left(a+\frac{1}{a}\right) = 21-3=18$.
Then $a^3+1+\frac{1}{a^3}=19$. Myltiplying by $b^3=1$, $\Rightarrow$ $x^6+x^3y^3+y^6=19$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Needs to show Monotone While doing one of my works I need to show this expression is monotone(increasing).
$$e(2n+2)!\sum_{k=2n+2}^{\infty} \frac{(-1)^k}{k!}$$
I am sure this is increasing on $n$ because I explicitly computed for $n=1$ to $10$. And observed it is increasing but I need to prove it.
I tried many process such as diving by another consecutive term and trying to get more than or equal to $1$. Next I tried by taking the difference. None of my process worked here.
If anyone could show this is increasing then I would be very grateful.
| Let
$$
a_n
=(2n)! \sum_{k=2n}^{\infty} \frac{(-1)^k}{k!}
=(2n)! \sum_{k=n}^{\infty} \left(\frac{1}{(2k)!} - \frac{1}{(2k+1)!} \right)
$$
Clearly, $a_n > 0$ and
\begin{align}
\frac{a_n}{(2n)!} = \frac{2n}{(2n+1)!} + \frac{a_{n+1}}{(2n+2)!}
\end{align}
So
\begin{align}
a_n =\frac{2n}{2n+1} + \frac{a_{n+1}}{(2n+2)(2n+1)} \tag{1}
\end{align}
In particular
\begin{align}
a_n > \frac{2n}{2n+1} \tag{2}
\end{align}
It follows that
\begin{align}
a_{n+1} - a_n
&= -\frac{2n}{2n+1} + \left( 1 - \frac{1}{(2n+2)(2n+1)} \right)a_{n+1} \\\\
&> -\frac{2n}{2n+1} + \left( 1 - \frac{1}{(2n+2)(2n+1)}\right) \frac{2n+2}{2n+3} &\hbox{by $(2)$} \\\\
&= \frac{(2n+2)(2n+1)}{(2n+3)(2n+1)} - \frac{(2n)(2n+3)+1}{(2n+3)(2n+1)} \\\\
&= \frac{1}{(2n+3)(2n+1)} > 0
\end{align}
So the sequence $\{a_n\}$ is increasing as desired.
| {
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Power series method to solve this ODE: what did I do wrong? Problem: Use the power series method to solve the ODE $$ 2(x - 1) y' = 3y. $$
Attempt: I solved this with power series method, and then compared with the technique of separation of variables, and I'm not getting the same answer. Here is what I did:
We look for a power series around $x = 0$. Hence we look for a solution of the form $y(x) = \sum_{n=0}^{\infty} a_n x^n$. Then $y'(x) = \sum_{n=1}^{\infty} n a_n x^{n-1} $, and if we plug this in the ODE we get \begin{align*} 2(x-1) \sum_{n=1}^{\infty} n a_n x^{n-1} - 3 \sum_{n=0}^{\infty} a_n x^n = 0
\end{align*} or, after distributing and equating powers: \begin{align*} 2 \sum_{n=1}^{\infty} n a_n x^n - 2 \sum_{n=0}^{\infty} (n+1) a_{n+1} x^n - 3 \sum_{n=0}^{\infty} a_n x^n = 0 \end{align*} Hence by the identity theorem we get \begin{align*} \begin{cases} -2 a_1 - 3 a_0 &= 0 \qquad (n = 0) \\ 2n a_n - 2 (n+1) a_{n+1} - 3 a_n &= 0 \qquad (n \geq 1) \end{cases} \end{align*} From the first equation, we have $a_1 = - \frac{3}{2} a_0$, and from the second $$ a_{n+1} = a_n \frac{(2n-3)}{2(n+1)} $$ Since $$ a_2 = \frac{(-3)(-3)}{2 \cdot 2} a_0 $$ and $$ a_3 = \frac{(-3)(-3)(-3)}{2 \cdot 2 \cdot 2} a_0 $$ the general pattern is $$ a_n = (-1)^n (\frac{3}{2})^n a_0 $$ Hence the solution is given as $$ y(x) = a_0 \sum_{n=0}^{\infty} (- \frac{3}{2})^n x^n $$ and the radius of convergence is $R = 2 / 3$ because $\lim_{n \to \infty} \big( ( 3/2)^n \big)^{1/n} = 3/2 $ by Cauchy's test. I rewrote the solution as $$ y(x) = a_0 \bigg(\frac{1}{1 + \frac{3}{2} x} \bigg) = a_0 \big( \frac{2}{2 + 3x} \big). $$
But If I solve the ODE by seperation of variables, I get a different answer which is also what Maple gives me: $y(x) = C (x-1)^{3/2}$. So what did I do wrong here?
| There is a mistake in your computation of the coefficients $a_n$ (see below) :
One can reconize the binomial coefficients
\begin{pmatrix}
\frac{3}{2} \\
n
\end{pmatrix}
wich allows to link with $(1-x)^{3/2}$
| {
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Integration of $\frac{1}{(1+x^4)^\frac{1}{4}}$ This question has been puzzling me a lot. This is in the middle list of question(meaning its moderately tough). I have tried everything i could think of.
My try: $$\int \frac{1}{x(1+\frac{1}{x^4})^\frac{1}{4}}dx$$
$$-\frac{1}{4}\int \frac{-4x^4}{x^5(1+ \frac{1}{x^4})^\frac{1}{4}}dx$$
putting $$1 + \frac{1}{x^4} = t \iff x^4=\frac{1}{t-1} \iff dt = -\frac{4}{x^5}dx$$ and substituting in the integral,
$$-\frac{1}{4}\int \frac{1}{(t-1)t^\frac{1}{4}}dt$$
And Now I am stuck here.
| Try this let $t^4 = 1 + x^{-4}$
| {
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Prove by induction: $\left(\frac{1}{n+1}+\frac{1}{n+2}+...\frac{1}{2n}\right)^2<\frac{1}{2},n\ge 1$ Prove by induction: $\left(\frac{1}{n+1}+\frac{1}{n+2}+...\frac{1}{2n}\right)^2<\frac{1}{2},n\ge 1$
For $n=1$ inequality holds.
For $n=k$
$$\left(\frac{1}{k+1}+\frac{1}{k+2}+...\frac{1}{2k}\right)^2<\frac{1}{2}$$
For $n=k+1$
$$\left(\frac{1}{k+2}+\frac{1}{k+3}+...\frac{1}{2k+2}\right)^2<\frac{1}{2}$$
I don't know how to prove this.
Is it better to use direct method or contradiction?
| I have Cauchy-Schwarz inequality prove your inequality
we have
\begin{align*}\dfrac{1}{n+1}+\dfrac{1}{n+2}+\cdots+\dfrac{1}{n+n}&\le\sqrt{n\left(\dfrac{1}{(n+1)^2}+\dfrac{1}{(n+2)^2}+\cdots+\dfrac{1}{(n+n)^2}\right)}\\
&<\sqrt{n\left(\dfrac{1}{n(n+1)}+\dfrac{1}{(n+1)(n+2)}+\cdots+\dfrac{1}{(n+n-1)(n+n)}\right)}\\
&=\sqrt{n\left(\dfrac{1}{n}-\dfrac{1}{2n}\right)}\\
&=\dfrac{\sqrt{2}}{2}
\end{align*}
EDIT If you must use induction, I think you must induction this following Stronger inequality
$$\dfrac{1}{n+1}+\dfrac{1}{n+2}+\cdots+\dfrac{1}{n+n}\le\dfrac{\sqrt{2}}{2}-\dfrac{1}{4n+1}$$
because we only prove this
$$\dfrac{\sqrt{2}}{2}-\dfrac{1}{4n+1}-\dfrac{1}{n+1}+\dfrac{1}{2n+1}+\dfrac{1}{2n+2}<\dfrac{\sqrt{2}}{2}-\dfrac{1}{4n+5}$$
we only prove
$$\dfrac{1}{4n+5}+\dfrac{1}{2n+1}<\dfrac{1}{2n+2}+\dfrac{1}{4n+1}$$
$$\Longleftrightarrow (4n+5)(2n+1)(4n+1)+(4n+5)(2n+1)(2n+2)-(4n+5)(2n+2)(4n+1)-(2n+1)(2n+2)(4n+1)>0$$
$$\Longleftrightarrow 3>0$$
It it clear
| {
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Which of the constants A,B,C,D does T depend on? Let $f(x)=cos(5x)+Acos(4x)+Bcos(3x)+Ccos(2x)+Dcos(x)+E$ and $T=f(0)-f(\pi/5)+f(2\pi/5)-f(3\pi/5)+..-f(9\pi/5)$.Then out of A,B,C,D which does T depend on?
Hints please!
P.S:KVPY 2011 question
| Given $$f(x) = \cos 5x+A\cos 4x+B\cos 3x+C\cos 2x+d\cos x+E$$
Using $$f(\pi-x) = f(\pi+x)\Rightarrow f(x) = f(2\pi-x).$$
So we get $$\displaystyle f\left(\frac{\pi}{5}\right)=f\left(\frac{9\pi}{5}\right)\;\;\;,\;\;\; \displaystyle f\left(\frac{2\pi}{5}\right)=f\left(\frac{8\pi}{5}\right)$$
and $$\displaystyle f\left(\frac{3\pi}{5}\right)=f\left(\frac{7\pi}{5}\right)\;\;\;,\;\;\; \displaystyle f\left(\frac{4\pi}{5}\right)=f\left(\frac{6\pi}{5}\right)$$
Now $$T=f(0)-2\left[\displaystyle f\left(\frac{\pi}{5}\right)+f\left(\frac{3\pi}{5}\right)\right]+2\left[\displaystyle f\left(\frac{2\pi}{5}\right)+f\left(\frac{4\pi}{5}\right)\right]-f(\pi)$$
Now $$f(0)-f(\pi) = 2\left[...\right]$$
and $$f\left(\frac{\pi}{5}\right)+f\left(\frac{3\pi}{5}\right) = 2\left[...\right]$$
and $$f\left(\frac{2\pi}{5}\right)+f\left(\frac{4\pi}{5}\right) = 2\left[...\right]$$
| {
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How to solve $\lvert{x}\rvert - \lvert{2+x}\rvert = x$? How do I solve the following equation?
$$\lvert{x}\rvert- \lvert{2+x}\rvert= x$$
I was thinking about dividing it into 4 cases: plus plus, plus minus, minus plus and minus minus. What is the best way to solve this?
| Another way is to use $|x| = \sqrt{x^2}$. If $|x| - |2+x| = x$ then
$$\begin{align*}
\sqrt{x^2} - \sqrt{(x+2)^2} &= x\\
x^2 + (x+2)^2 -2\sqrt{x^2}\sqrt{(x+2)^2} &= x^2\\
(x+2)^2 &= 2\sqrt{x^2} \sqrt{(x+2)^2}\\
(x+2)^4 &= 4x^2(x+2)^2
\end{align*}$$
So either $x= -2$, which doesn't satisfy the original equation, or else
$$(x+2)^2 = 4x^2$$
which is a quadratic equation in $x$. This has the roots $x=2$ and $x=-2/3$. Now you just have to check whether either of these satisfies the original equation, and only $x=-2/3$ does, so this is the only solution.
| {
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How should I go about solving this system of equations using Gauss Elimination? $${ x }_{ 1 }+2{ x }_{ 2 }-{ x }_{ 3 }+2{ x }_{ 4 }=0\\ { x }_{ 2 }+{ x }_{ 3 }-2{ x }_{ 4 }+2{ x }_{ 5 }=0\\ 2{ x }_{ 1 }+{ x }_{ 2 }-5{ x }_{ 3 }-4{ x }_{ 5 }=0$$
Steps I took:
$$\left[\begin{array}{rrrrr|r}
1 & 2 & -1 & 2 & 0 & 0 \\
0 & 1 & 1 & -2 & 2 & 0 \\
2 & 1 & -5 & 0 & -4 & 0
\end{array}\right]$$
$$\Rightarrow { R }_{ 3 }={ R }_{ 3 }-2{ R }_{ 1 } =$$
$$\left[\begin{array}{rrrrr|r}
1 & 2 & -1 & 2 & 0 & 0 \\
0 & 1 & 1 & -2 & 2 & 0 \\
0 & -3 & -2 & -6 & -4 & 0
\end{array}\right]$$
$$\Rightarrow { R }_{ 2 }={ R }_{ 3 }-2{ R }_{ 2 } =$$
$$\left[\begin{array}{rrrrr|r}
1 & 2 & -1 & 2 & 0 & 0 \\
0 & -1 & 0 & 10 & 0 & 0 \\
0 & -3 & -2 & -6 & -4 & 0
\end{array}\right]$$
$$\Rightarrow { R }_{ 3 }={ R }_{ 3 }-3{ R }_{ 2 } =$$
$$\left[\begin{array}{rrrrr|r}
1 & 2 & -1 & 2 & 0 & 0 \\
0 & -1 & 0 & -10 & 0 & 0 \\
0 & 0 & -2 & 24 & -4 & 0
\end{array}\right]$$
Where do I go from here in order to find the solutions for this system? I can already tell that I ma going to have an infinite amount of solutions because I have more unknown than constraints. I just don't know how to (properly) apply Gauss Elimination to this. A hint in the right direction would help a lot.
| multiplying the first equation by $-2$ and adding to the third equation we get
$$x_1+2x_2-x_3+3x_4=0$$
$$x_2+x_3+2x_4+2x_5=0$$
$$-3x_2-2x_3-x_4-4x_5=0$$
multiplying the second equation by $3$ and adding to the third we obtain
$$x_2+2x_5=0$$
| {
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The sum of non real roots of the polynomial equation $x^3+3x^2+3x+3=0$ Problem :
The sum of non real roots of the polynomial equation $x^3+3x^2+3x+3=0$
(a) equals 0
(b) lies between 0 and 1
(c)lies between -1 and 0
(d) has absolute value bigger than 1
My approach :
The discriminant of cubic equation $ax^3+bx^2+cx+d=0$ is given by
$\Delta = 18abcd -4b^3d +b^2c^2 -4ac^3 -27a^2d^2$
$\Delta = -108 < 0 $ Therefore the equation has one real and two non real roots.
But how to find the roots of this equation not getting any idea please help . thanks
| Hint: The equation can be written as
$$(x+1)^3=-2.$$
Hence the three roots are $\sqrt[3]2-1$, $\sqrt[3]2\omega-1$, $-\sqrt[3]2\omega-1$, where $\omega$ is a cubic root of unit.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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if $a(b-c)x^2+b(c-a)x+c(a-b)=0$ has repeated roots prove... if the equation $$a(b-c)x^2+b(c-a)x+c(a-b)=0$$ has repeated roots prove that $${1\over a}, {1\over b},{1\over c} $$ are in Arithmetic Progression
Any idea about how to go about solving this ? Thanks is advance!
| With $a,b,c\ne 0$, observe that $1/a,1/b,1/c$ is an Arithmetic Progression iff $$K=0\text {, where}$$ $$K=a b+b c-2a c.$$ The condition that the quadratic has a repeated root is that the discriminant is zero. We have $$0=b^2(c-a)^2-4a c(a-b)(b-c)\iff$$ $$0=b^2c^2+b^2a^2-2b^2a c-4a c(-b^2+b c+b a-ac)\iff$$ $$0=b^2c^2+b^2a^2+2a c b^2-4a c(b c+b a-a c)\iff$$ $$0= b^2c^2+b^2a^2+2a c b^2-4 a c(b c+b a-2a c)-4a^2c^2\iff$$ $$0=b^2c^2+b^2a^2+2a c b^2-4a c K-4a^2c^2\iff$$ $$0=(b c+b a)^2-4a c K-4a^2c^2\iff$$ $$0=(b c+b a)^2 -(2a c)^2-4a c K\iff$$ $$0=(b c+b a-2a c)(b c+b a+2a c)-4a c K$$ $$0=K(b c+b a+2a c)-4a c K\iff$$ $$0=K(b c+b a-2a c)\iff$$ $$0=K^2.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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roots of the quadratic equation $(a^4+b^4)x^2+4abcdx+(c^4+d^4)=0$ are real
If $a,b,c,d\in \mathbb{R}$ and roots of the quadratic equation $(a^4+b^4)x^2+4abcdx+(c^4+d^4)=0$
are real.Then prove that roots are equal.
$\bf{My\; Try::}$ Given $(a^4+b^4)x^2+4abcdx+(c^4+d^4)=0\;$ Then we can write it as
$$\left[(a^2x)^2+c^4-2a^2c^2x+(b^2x)^2+d^4-2b^2d^2x+2a^2c^2x+2b^2d^2x+4abcdx\right]=0$$
So $$(a^2x-c^2)^2+(b^2x-d^2)^2+2x(ac+bd)^2=0$$
Now I did not understand How can I proceed further.
Although I have a knowledge of $\bf{Discriminant\; Method.}$
So plz explain me above method which i am trying above.
Thanks.
| The discriminant is non-positive because of the geometric-quadratic means inequality. Indeed the reduced discriminant is
$$\Delta'=4a^2b^2c^2d^2-(a^4+b^4)(c^4+d^4)$$
Sert $A=a^2, B=b^2, \&c.$ We have to prove $\;4ABCD\ge (A^2+B^2)(C^2+D^2)$
Now by the GQM inequality, we have
\begin{alignat*}{2}\sqrt{AB}&\le\sqrt\frac{A^2+B^2}2, &\quad\sqrt{CD}&\le\sqrt\frac{C^2+D^2}2\\
\iff 2AB&\le A^2+B^2,&2CD & \le C^2+D^2
\end{alignat*}
so multiplying both sides of the inequalities, we get
$$4ABCD\le(A^2+B^2)( C^2+D^2) $$
Thus we have real roots if and only if $4ABCD=(A^2+B^2)( C^2+D^2) $, i. e. if there is a double root, which is necessarily equal to the half sum of the roots by Vieta's formulae:
$$-\frac{2abcd}{a^4+b^4}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove inequality $\left| \frac{x+yz}{x^2+y^2} \right| \leq 1$ for $x^2+y^2-z^2=1$ Prove this:
If $x,y,z \in \mathbb R$ and $x^2+y^2-z^2=1$, then $$\left| \frac{x+yz}{x^2+y^2} \right| \leq 1 $$ holds.
Own ideas:
If we eliminate $z$ the inequality is equivalent to $$\left| \frac{x \pm y\sqrt{x^2+y^2-1}}{x^2+y^2} \right| \leq 1.$$
| WLOG, let $z=\tan C$ and $x=\sec C\cos B,y=\sec C\sin B$
$$\dfrac{x+yz}{x^2+y^2}=\cos^2C(\sec C\cos B+\sec C\sin B\tan C)$$
$$=\cos C\cos B+\sin B\sin C=\cos(B-C)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the least squares solution of the linear system...... Hi I'm looking for the least squares solutions of....
$$ \begin{pmatrix} 1&-1 \\ -1&2 \\ -1&0 \end{pmatrix} \begin{pmatrix} x\\y \end{pmatrix} = \begin{pmatrix} 3 \\ 4 \\ 5\end{pmatrix} $$
So Assuming this goes by $ A\vec{x}=\vec{b}$
Then using.... $A^T A \vec{x} = A^T \vec{b} $
$$ \begin{pmatrix} 1&-1&-1 \\ -1&2&0 \end{pmatrix} \begin{pmatrix} 1&-1 \\ -1&2 \\ -1&0 \end{pmatrix} \vec{x}= \begin{pmatrix} 1&-1&-1 \\ -1&2&0 \end{pmatrix} \begin{pmatrix} 3 \\ 4 \\ 5\end{pmatrix} $$
Which solves to be
$$ \begin{pmatrix} 3&-3 \\ -3&5 \end{pmatrix} \vec{x} = \begin{pmatrix} -6 \\ 5 \end{pmatrix} $$
which i solve to get
$$ \vec{x} = \begin{pmatrix} -7/3 \\ 3/2 \end{pmatrix} $$
| Since you are only asking for the solution...
$$ \left( \begin{vmatrix} 1 & -1 \\ -1 & 2 \\ -1 & 0 \end{vmatrix}^\intercal \begin{vmatrix} 1 & -1 \\ -1 & 2 \\ -1 & 0 \end{vmatrix} \right) \begin{pmatrix} x \\ y \end{pmatrix} = \left( \begin{vmatrix} 1 & -1 \\ -1 & 2 \\ -1 & 0 \end{vmatrix}^\intercal \begin{pmatrix} 3 \\ 4 \\ 5 \end{pmatrix} \right) $$
$$ \begin{vmatrix} 3 & -3 \\ -3 & 5 \end{vmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} -6 \\ 5 \end{pmatrix} $$
$$ \begin{pmatrix} x = -\frac{5}{2} \\ y = -\frac{1}{2} \end{pmatrix} $$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to compute this gross limit.
How do I compute this limit?
$$
\lim_{n \to \infty}
\frac{\left(1 + \frac{1}{n} + \frac{1}{n^2}\right)^n -
\left(1 + \frac{1}{n} - \frac{1}{n^2}\right)^n
}{
2 \left(1 + \frac{1}{n} + \frac{1}{n^2}\right)^n -
\left(1 + \frac{1}{n} - \frac{1}{n^2 + 1}\right)^n -
\left(1 + \frac{1}{n} - \frac{1}{n^2 (n^2 +1)}\right)^n
}
$$
I think I got the correct limit by using fast converging limits to $e$.
In particular I used truncated Taylor series for the sqrt and 4th root.
Or squares and bisquares.
Example
$(1+1/2n)^{2n}$
Becomes
$(1 + 1/n + 1/4n^2)^n.$
In combination with l'hopital it gives me the answer.
But I guess that is not a very good (fast) method.
| You might get some easy simplifications by extending the fraction with the factor $(1-\frac1n)^n$ so that all terms transform to the form
$$
\left(1+\frac{c_n}{n^2}\right)^n=1+\frac{c_n}{n}+O(n^{-2}).
$$
to obtain
\begin{align}
&\ \lim_{n \to \infty}
\frac{\left(1 - \frac{1}{n^3}\right)^n -
\left(1 - \frac{2}{n^2} + \frac{1}{n^3}\right)^n
}{
2 \left(1 - \frac{1}{n^3}\right)^n -
\left(1 - \frac{2n^2+1-n}{n^2 (n^2 + 1)}\right)^n -
\left(1 - \frac{n^2+2}{n^2 (n^2+1)} + \frac{1}{n^3 (n^2 +1)}\right)^n
}
\\ \\
&=\lim_{n \to \infty}
\frac{\left(1 + O(n^{-2})\right) -
\left(1-\frac2n + O(n^{-2}) \right)
}{
2 \left(1 + O(n^{-2})\right) -
\left(1 - \frac{2}{n} + O(n^{-2})\right) -
\left(1 - \frac{1}{n} + O(n^{-2})\right)
}
\\ \\
&=\frac 23
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Greatest Common Divisor Proof: $\gcd(m^2-n^2, m^2+n^2) = 1$ Prove that if $\gcd(m,n) = 1$ and $m+n \equiv 1 (\text{mod} ~2)$ then $\gcd(m^2-n^2, m^2+n^2) = 1$
Workings:
Suppose that $\gcd(m,n) = 1$ and $m+n \equiv 1 (\text{mod} ~2)$
$\gcd(m^2-n^2, m^2+n^2)$
$= gcd((m-n)(m+n), (m-n)(m+n)+2n^2)$
Now I know that $m+n=1 (\text{mod} ~2)$ means that one of $m$ or $n$ is odd or even
But now I'm not too sure on what to do.
Any help will be appreciated.
| I like to write it this way: $p$ is assumed to be a prime that divides both $m^2 - n^2$ and $m^2 + n^2.$ Then $p$ divides their sum, so $p | 2 m^2$ and $p|2m.$ Either $p=2$ or $p |m.$
Also $p$ divides their difference, so $p | 2 n^2$ and $p|2n.$ Either $p=2$ or $p |n.$
If $p \neq 2,$ then $p$ divides both $m,n.$ We know this is a contradiction because $\gcd(m,n) = 1.$
Finally, try $p=2.$ This fails as well, because $m+n,$ and so $m-n,$ are odd, thence $m^2 - n^2$ is odd and not divisible by $2.$
There you have it, we have contradicted the assumption that there is any prime $p$ that divides both $m^2 - n^2$ and $m^2 + n^2,$ therefore $\gcd(m^2 - n^2, m^2 + n^2) = 1.$
| {
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"timestamp": "2023-03-29T00:00:00",
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Computing Two-Norm for interpolation of functions $$ f(x) = x^3, p(x) = (3/2)x^2 − (1/2)x $$
The two-norm of f(x) - p(x) is: $$( \int_0^1 (f(x) - p(x))^2 dx )^{1/2} $$
p(x) interpolates f(x) at $$x=0, x=1/2, x=1$$
The result of the two-norm computation should be $$\sqrt{210}/420$$
I can't seem to calculate the numerator $\sqrt{210}$ correctly. What would be the proper way?
| I suppose some minor mistakes (probably a sign error in the development of the square).
If $$f(x)=x^3 \quad , \quad p(x)=\frac{3 }{2}x^2-\frac{1}{2}x$$ then $$f(x)-p(x)=x^3-\frac{3 }{2}x^2+\frac{1}{2}x$$ Now square carefuly and group terms for same powers of $x$ to get $$\big(f(x)-p(x)\big)^2=x^6-3 x^5+\frac{13 }{4}x^4-\frac{3 }{2}x^3+\frac{1}{4}x^2$$ Integrate each term $$\int \big(f(x)-p(x)\big)^2\,dx=\frac{1}{7}x^7-\frac{1}{2}x^6+\frac{13 }{20}x^5-\frac{3 }{8}x^4+\frac{1}{12}x^3$$
I am sure that you can take from here and get your result.
| {
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Show that if $a\neq b$ then $a^3+a\neq b^3+b$ Show that if $a\neq b$ then $a^3+a\neq b^3+b$
We assume that $a^3+a=b^3+b$ to show that $a=b$
$$\begin{align}
a^3+a=b^3+b &\iff a^3-b^3=b-a\\
&\iff(a-b)(a^2+ab+b^2)=b-a\\
&\iff a^2+ab+b^2=-1
\end{align}$$
Im stuck here !
| $$a^3+a=b^3+b\iff a^3-b^3=b-a$$
$$\iff (a-b)\left(a^2+ab+b^2\right)=b-a$$
If $a=b$, then we're done. For contradiction, assume $a\neq b$. Then $a-b\neq 0$ and we can divide both sides by $a-b$:
$$a^2+ab+b^2=-1\iff 4a^2+4ab+4b^2=-4$$
$$\iff (2a+b)^2+3b^2=-4,$$
contradiction, because $(2a+b)^2+3b^2\ge 0$ for all $a,b\in\Bbb R$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find all the real solutions to the equation: $(x+i)^n-(x-i)^n=0$ Find all the real solutions to the equation: $$(x+i)^n-(x-i)^n=0$$
The answer is given, I will type it out until the line which is unclear to me (meaning I understand all the steps leading up to the last line).
$$\left(\frac{x+i}{x-i}\right)^n=1 \implies \frac{x+i}{x-i}=\sqrt[n]{1}=\cos \frac{2k \pi}{n}+i\sin\frac{2k\pi}{n},\qquad k=0,1,\ldots,(n-1)$$
Now we find $x$:
$$x=\frac{\sin\frac{2k\pi}{n}-i(1+\cos\frac{2k\pi}{n})}{1-\cos\frac{2k\pi}{n}-i\sin\frac{2k\pi}{n}}$$
The rest is clear, just this last line.
| I think that you could go one step further $$\frac{x+i}{x-i}= a + i b \implies x=i\frac{a+i b+1}{a+i b-1}=i\frac{a+i b+1}{a+i b-1}\times\frac{a-i b-1}{a-i b-1}$$ Develop and isolate the real and imaginary parts to get $$x=\frac{2 b}{(a-1)^2+b^2}+ i\frac{a^2+b^2-1}{(a-1)^2+b^2}$$ But, an here starts the beauty in your problem : $a$ is a cosine and $b$ is a sine which makes the imaginary part disappearing since $a^2+b^2=1$.
So what is left is $$x=\frac{2 b}{(a-1)^2+b^2}=\frac{2 b}{a^2-2a+1+b^2}=\frac{2 b}{2-2a}=\frac{ b}{1-a}=\frac{ \sin(\theta)}{1-\cos(\theta)}$$ Now, use the half angle formula $$x=\frac{ \sin(\theta)}{1-\cos(\theta)}=\frac {2 \sin(\frac \theta 2)\cos(\frac \theta 2)}{2\sin^2(\frac \theta 2)}=\cot (\frac{\theta}{2})$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Help with summation: $\sum_{k=1}^\infty\frac{k(k+2)}{15^k}$ How can one evaluate the below sum? Any help would be greatly appreciated.
$$\sum_{k=1}^\infty\frac{k(k+2)}{15^k}$$
| $$\sum_{k=1}^\infty\frac{k(k+2)}{15^k}=\sum_{k=1}^\infty\frac{k(k+1)}{15^k}+\sum_{k=1}^\infty\frac{k}{15^k}$$
start with the first term
depending on the geometric series
$$\frac{1}{1-x}=\sum_{k=0}^{\infty }x^k$$
$$\frac{d}{dx}(\frac{1}{1-x})=\sum_{k=1}^{\infty }kx^{k-1}$$
$$\frac{d^2}{dx^2}(\frac{1}{1-x})=\sum_{k=2}^{\infty }k(k-1)x^{k-2}$$
$$=\sum_{k=1}^{\infty }k(k+1)x^{k-1}=\frac{1}{x}\sum_{k=1}^{\infty }k(k+1)x^{k}$$
$$x\frac{d^2}{dx^2}(\frac{1}{1-x})=\sum_{k=1}^{\infty }k(k+1)x^{k}$$
now, for second term
$$\frac{d}{dx}(\frac{1}{1-x})=\sum_{k=1}^{\infty }kx^{k-1}$$
$$x\frac{d}{dx}(\frac{1}{1-x})=\sum_{k=1}^{\infty }kx^{k}$$
$$\sum_{k=1}^\infty\frac{k(k+2)}{15^k}=x[\frac{d^2}{dx^2}(\frac{1}{1-x})+\frac{d}{dx}(\frac{1}{1-x})]$$
then plug $x=\frac{1}{15}$ to get what you need
| {
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"timestamp": "2023-03-29T00:00:00",
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Use integration by parts to find the integral $\int\frac{\sqrt {4x^2-9}}{x^2}dx$ $$\int\frac{\sqrt {4x^2-9}}{x^2}dx$$
I tried to solve this using integration by parts, but I come up with something that is much more difficult to solve. How can this be solved?
| By substitution: $\DeclareMathOperator\ach{arg\,cosh}$
Set $ x=\dfrac32\cosh t,\enspace t\ge0$, whence $\mathrm d\mkern1mu x=\dfrac32\sinh t\,\mathrm d\mkern1mu t$. With some hyperbolic trigonometry, we get
\begin{align*}
\int \frac{\sqrt{4x^2-9}}{x^2}\,\mathrm d\mkern1mu x&=2\int \frac{\sinh t}{\cosh^2t}\sinh t\,\mathrm d\mkern1mu t =2\int \tanh^2t\,\mathrm d\mkern1mu t \\&=2\int\bigl(1-(1-\tanh^2t)\bigr)\,\mathrm d\mkern1mu t=2(t-\tanh t)\\
&=2\Biggl(\ach\frac{2x}3-\frac{\sinh\bigl(\ach\frac{2x}3\bigr)}{\frac{2x}3}\Biggr)\\
&=2\Biggl(\ach\frac{2x}3-\frac{3\sqrt{\frac{4x^2}9-1}}{2x}\Biggr)=2\Biggl(\ach\frac{2x}3-\frac{\sqrt{4x^2-9}}{2x}\Biggr)\\
&=2\ln\bigl(2x+\sqrt{4x^2-9}\bigr)-\frac{\sqrt{4x^2-9}}{x}+\text{constant}.
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
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Taylor Series and Differentiation with Sigma notation $f(x) = \frac{x}{(2-3x)^2}$ Use Term By Term Differentiation to Find the Taylor Series about $x$=3 for
Give The Open Interval of Convergence and express as sigma notation
$\sum A_n(x-3)^n$
$f(x) = \frac{x}{(2-3x)^2}$
So I have Found the Taylor Series for 1/(2-3x) to be
$\sum{(-3)^{n}\cdot(x-3)^{n} }\cdot{(-7)^ {n+1}}$
How Do you find the original function taylor series and its interval of convergence and then express in sigma notation of the form $\sum A_n(x-3)^n$
| you can make a change of variable $x = 3 + h.$ then we have
$$\begin{align} f(x) &= \frac{x}{(2-3x)^2} = \frac{3+h}{(2 - 9-3h)^2}=\frac 1{49}(3+h)\left(1 + \frac{3h}7\right)^{-2} \\
&=\frac 1{49}(3+h)\left(1 -\frac 21 \frac{3h}7 + \frac{2\cdot3}{1\cdot2}\left(\frac{3h}{7}\right)^2 - \frac{2\cdot3 \cdot 4}{1\cdot2\cdot 3}\left(\frac{3h}{7}\right)^2 + \ldots\right) \end{align} $$
i will let you multiply out the two factors and collect the like terms.
| {
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"timestamp": "2023-03-29T00:00:00",
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How many solutions this a multidimensional system have? How many solutions have this system of equations:
\begin{array}{l}
\ 3x^2+y+2xy^2-3 = 0 \\
\ x + 2yx = 0
\end{array}
The more general case is when we want to solve two nonlinear algebraic equations for two unknowns
\begin{array}{l}
\ {A_1}{x^2} + {B_1}xy + {C_1}{y^2} + {D_1}x + {E_1}y + {F_1} = 0 \\
\ {A_2}{x^2} + {B_2}xy + {C_2}{y^2} + {D_2}x + {E_2}y + {F_2} = 0
\end{array}
| There are two roots of the second equation.
$x=0$ substituted into the first equation gives $ y=3.$
$y=-\frac12$ substituted into the first equation gives two solutions from the quadratic resulting in:
$$ x = \dfrac{-3 \pm \sqrt{82}}{2} $$
So, in all 3 different roots and 3 solutions.
| {
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"timestamp": "2023-03-29T00:00:00",
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If $ \frac{y}{x-z} = \frac{x+y}{z} = \frac{x}{y} $ find the ratio of $x$, $y$ and $ z$ I have this question from higher algebra by Hall and Knight:
if $ \frac{y}{x-z} = \frac{x+y}{z} = \frac{x}{y} $ then find the ratio of x,y and z?
There are two answers given for this question, the first is $\frac x4 =\frac y2 =\frac z3$ and the second is $\frac x1 =\frac y{-1} =\frac z0 $. Now I solve this question in the following manner:
Adding numerator and denominator gives $ \frac{y}{x-z} = \frac{x+y}{z} = \frac{x}{y}= 2\frac{x+y}{x+y} =2$ (when x+y is not zero).This gives the first answer.
When (x+y ) is zero => $ y= -x $ that is $\frac xy = -1$ and now there are two things
(i)If I put these values in the original expression I get $ \frac{y}{x-z} =\frac 0z= -1$ this implies 0= 1 where am I making the mistake?
(ii)Also from the original expression I have$ \frac{x+y}{z} = \frac{x}{y}$ multiplying by z I get $ x+y =z\frac xy$. This implies that z=0 and x:y:z =x:-x:0 = 1:-1:0 and this gives the second answer.But the problem is how z can be zero when it appears in the denominator in the expression
I think these things are very basic but still I am stuck.Could anyone please help me in knowing where the mistake is?
| y/(x-z)=(y+x)/z=x/y = be k
y= kx - kz ..... 1
y+x = kz ...... 2
x=yk ................3
Add 1 & 2 , 2y+ x = kx , substitute 3 , 2y+ yk = k *yk
2 +k = k² , k²- k - 2 = 0 , k= 2 or k = -1
for k = 2 , x = 2y , substitute in 2 , 3y = 2z
x:y:z = 2y:y:3y/2 = 4:2: 3
| {
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Where did I mistake to integrate $I=\int\sqrt{\frac{\sin(x-\alpha)}{\sin(x+\alpha)}} \; dx\; ?$ It was given to integrate $$I=\int\sqrt{\frac{\sin(x-\alpha)}{\sin(x+\alpha)}} \; dx.$$
Attempt:
\begin{align}I&= \int\sqrt{\frac{\sin((x+\alpha)-2\alpha)}{\sin(x+\alpha)}}\; dx\\&= \sqrt{\sin 2\alpha}\int\sqrt{\cot 2\alpha - \cot (x+\alpha)}\;dx\; .\end{align}
Then I took \begin{align}\cot 2\alpha - \cot (x+\alpha)= z^2 \\ \implies [1 + \cot^2(x+\alpha)]\;dx= 2z\;dz\\ \implies [1+\{\cot 2\alpha - z^2\}^2]\; dx= 2z\;dz\; .\end{align}
Now,
\begin{align}\int\sqrt{\cot 2\alpha - \cot (x+\alpha)}\; dx\\ &= \int\frac{z\cdot 2z}{1+\{\cot 2\alpha - z^2\}^2}\;dz\\ &= \int\frac{2}{\dfrac{(1+ \cot^2 2\alpha)}{z^2} + z^2 -2\cot2\alpha}\; dz\\ &= \int \frac{\left(1+ \dfrac{\sqrt{(1+ \cot^2 2\alpha)}}{z^2}\right)+ \left(1- \dfrac{\sqrt{(1+ \cot^2 2\alpha)}}{z^2}\right)}{\dfrac{(1+ \cot^2 2\alpha)}{z^2} + z^2 -2\cot2\alpha}\; dz \; .\end{align}
This makes me to break the integral to get \begin{align}\int\sqrt{\cot 2\alpha - \cot (x+\alpha)}\; dx\\ &= \int \frac{\left(1+ \dfrac{\sqrt{(1+ \cot^2 2\alpha)}}{z^2}\right)+ \left(1- \dfrac{\sqrt{(1+ \cot^2 2\alpha)}}{z^2}\right)}{\dfrac{(1+ \cot^2 2\alpha)}{z^2} + z^2 -2\cot2\alpha}\; dz\\ \\ &= \int\frac{\left(1+ \dfrac{\sqrt{(1+ \cot^2 2\alpha)}}{z^2}\right)}{\left(z-\dfrac{\sqrt{(1+ \cot^2 2\alpha)}}{z}\right) +\text{const.}}\; dz +\int\frac{\left(1- \dfrac{\sqrt{(1+ \cot^2 2\alpha)}}{z^2}\right)}{\left(z+\dfrac{\sqrt{(1+ \cot^2 2\alpha)}}{z}\right) +\text{const.}}\; dz \\ &=-\frac{1}{z- \dfrac{(1+ \cot^2 2\alpha)}{z}}-\frac{1}{z+ \dfrac{(1+ \cot^2 2\alpha)}{z}} .\end{align}
So, $$I= -\sqrt{\sin 2\alpha}\left(\frac{1}{z- \dfrac{(1+ \cot^2 2\alpha)}{z}}+\frac{1}{z+ \dfrac{(1+ \cot^2 2\alpha)}{z}}\right)\; .$$
Unfortunately, my book gave the answer as $$I= \cos\alpha \cos^{-1} (\cos x\sec\alpha) - \sin\alpha\log\left(\sin x+ \sqrt{\sin^2 x - \sin^2\alpha}\right)\; .$$
This is not near to my solution; I've made somewhere a big plunder.
So, could anyone please tell me where I did mistake? Is my attempted technique wrong?
| \begin{align}\int\sqrt{\cot 2\alpha - \cot (x+\alpha)}\; dx\\ &= \int \frac{\left(1+ \dfrac{\sqrt{(1+ \cot^2 2\alpha)}}{z^2}\right)+ \left(1- \dfrac{\sqrt{(1+ \cot^2 2\alpha)}}{z^2}\right)}{\dfrac{(1+ \cot^2 2\alpha)}{z^2} + z^2 -2\cot2\alpha}\; dz\\ \\ &= \int\frac{\left(1+ \dfrac{\sqrt{(1+ \cot^2 2\alpha)}}{z^2}\right)}{\left(z-\dfrac{\sqrt{(1+ \cot^2 2\alpha)}}{z}\right)^2 +2\tan\alpha}\; dz +\int\frac{\left(1- \dfrac{\sqrt{(1+ \cot^2 2\alpha)}}{z^2}\right)}{\left(z+\dfrac{\sqrt{(1+ \cot^2 2\alpha)}}{z}\right)^2 -2\cot\alpha}\; dz \\ &=\int\frac{du}{u^2+2\tan\alpha}+\int\frac{dv}{v^2-2\cot\alpha}\end{align}
where $$u=z-\frac{\sqrt{1+\cot^2 2\alpha}}{z}$$ and $$v=z+\frac{\sqrt{1+\cot^2 2\alpha}}{z}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1528793",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Calculate the limit $\lim_{n\to\infty} \left(\frac{n^2}{n^2 + 1} \right)^{(n-1)/(n+1)}$ $$\lim_{n\to\infty} \left(\frac{n^2}{n^2 + 1} \right)^{(n-1)/(n+1)}$$
My attempt
\begin{align}
\lim_{n\to\infty} \left(\frac{n^2}{n^2 + 1} \right)^{(n-1)/(n+1)} &= \lim_{n\to\infty} \left(\frac{n^2+1}{n^2} \right)^{(-n+1)/(n+1)} \\
&= \lim_{n\to\infty} \left(\frac{n^2+1}{n^2} \right)^{1 + (-2n)/(n+1)} \\
&= \lim_{n\to\infty} \left(\frac{n^2+1}{n^2} \right)\left(\frac{n^2+1}{n^2} \right)^{(-2n)/(n+1)}
\end{align}
since $\left(\frac{n^2+1}{n^2} \right) \to 1$ when $n\to\infty$
$$\lim_{n\to\infty} \left(\frac{n^2+1}{n^2} \right)\left(\frac{n^2+1}{n^2} \right)^{(-2n)/(n+1)} = \lim_{n\to\infty} \left(\frac{n^2+1}{n^2} \right)^{(-2n)/(n+1)}$$
Also I tried to make similar simplifications into the braсkets but nothing happens and no proof that limit $= 1$ or whatever.
And here is a rule for the task. This is a limit of a sequence so no usage of functional simplifications and derives are allowed. If you have really beautiful solution for the task then post it anyway.
| Given that $n^2/(n^2+1) < 1$, and $1>\frac{n-1}{n+1} >0$ for $n>1$, we can upper bound $\left(\frac{n^2}{n^2+1} \right)^{\frac{n-1}{n+1}} \leq 1^{\frac{n-1}{n+1}} \leq 1$. So, the limit is upper bounded by $1$.
Now, note that we have the lower bound $\left(\frac{n^2}{n^2+1} \right)^{\frac{n-1}{n+1}} \geq \frac{n^2}{n^2+1}$ since we have (something smaller than 1)^(something smaller than 1). The limit of $\frac{n^2}{n^2+1}$ is $1$. So the limit is lower bounded by $1$.
The upper and lower bounds match, so its limit is $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1529304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Fibonacci sequence, number of trees, probability How can I prove that the numbers of orderings described below are Fibonacci numbers?
We are given $p_1, \dots , p_n > 0$ such that each $p_i= \frac{1}{2^k}$ for $i \in \{1, \dots , n\}, \ \ k \in \mathbb{N}$ and $$p_1 + \dots +p_n = 1$$ and the sequence $(p_1, ..., p_n)$ must be non rising.
I've computed the numbers of such orderings for small $n$.
And for $n=1$ we have only one possibility: $(1)$ For $n=2$ we have $$\left(\frac{1}{2} , \frac{1}{2}\right)$$ so also one possibility. For $n=3$ we have one: $$\left( \frac{1}{2}, \frac{1}{4}, \frac{1}{4}\right)$$ for $ n=4$ I've found $2$: $$\left( \frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{8}\right) \ \ \left( \frac{1}{4}, \frac{1}{4}, \frac{1}{4}, \frac{1}{4}\right)$$ for $n=5$ we have 3: $$\left( \frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{16}, \frac{1}{16}\right) \ \ \left( \frac{1}{2}, \frac{1}{8}, \frac{1}{8}, \frac{1}{8}, \frac{1}{8}\right) \ \ \left( \frac{1}{4}, \frac{1}{4}, \frac{1}{4}, \frac{1}{8}, \frac{1}{8}\right)$$ for $n=6$ I have $5$:
$$\left(\frac{1}{2},\frac{1}{4}, \frac{1}{8}, \frac{1}{16}, \frac{1}{32}, \frac{1}{32} \right) \ \ \left(\frac{1}{2},\frac{1}{4}, \frac{1}{16}, \frac{1}{16}, \frac{1}{16}, \frac{1}{16} \right) \ \ \ \\ \left(\frac{1}{2},\frac{1}{8}, \frac{1}{8}, \frac{1}{8}, \frac{1}{16}, \frac{1}{16} \right) \ \ \left(\frac{1}{4},\frac{1}{4}, \frac{1}{4}, \frac{1}{8}, \frac{1}{16}, \frac{1}{16} \right), \left(\frac{1}{4},\frac{1}{4}, \frac{1}{8}, \frac{1}{8}, \frac{1}{8}, \frac{1}{8} \right) $$ and so on. But I don't know how to justify it mathematically, that means - how to find an occurence relation for different $n$s .
| It is not the Fibonacci sequence, as it starts $1, 1, 1, 2, 3, 5, 9, 16, \ldots $ so has too many examples when $n \ge 7$. For $7$ parts, the examples are:
*
*$ \frac1{2} + \frac1{4} + \frac1{8} + \frac1{16} + \frac1{32} + \frac1{64} + \frac1{64} $
*$ \frac1{2} + \frac1{4} + \frac1{8} + \frac1{32} + \frac1{32} + \frac1{32} + \frac1{32} $
*$ \frac1{2} + \frac1{4} + \frac1{16} + \frac1{16} + \frac1{16} + \frac1{32} + \frac1{32} $
*$ \frac1{2} + \frac1{8} + \frac1{8} + \frac1{8} + \frac1{16} + \frac1{32} + \frac1{32} $
*$ \frac1{2} + \frac1{8} + \frac1{8} + \frac1{16} + \frac1{16} + \frac1{16} + \frac1{16} $
*$ \frac1{4} + \frac1{4} + \frac1{4} + \frac1{8} + \frac1{16} + \frac1{32} + \frac1{32} $
*$ \frac1{4} + \frac1{4} + \frac1{4} + \frac1{16} + \frac1{16} + \frac1{16} + \frac1{16} $
*$ \frac1{4} + \frac1{4} + \frac1{8} + \frac1{8} + \frac1{8} + \frac1{16} + \frac1{16} $
*$ \frac1{4} + \frac1{8} + \frac1{8} + \frac1{8} + \frac1{8} + \frac1{8} + \frac1{8} $
The sequence is described by OEIS $A002572$ with several references.
Empirically it seems that number of examples with $n$ parts is slightly more than about $$0.141853\ldots \times 1.794147\ldots^n$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1530020",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Sum up trigonometric series $$\cos \frac{2π}{2013} +\cos \frac{4π}{2013} +\cdots+\cos \frac{2010π}{2013} + \cos \frac{2012π}{2013}$$
How to sum it up?
*Calculator is not allowed.
| Original Approach: Using the formula for the sum of a geometric series, we get
$$
\sum_{k=0}^{n-1}e^{\frac{2\pi i}nk}=\frac{e^{2\pi i}-e^0}{e^{\frac{2\pi i}n}-1}=0\tag{1}
$$
and using Euler's Formula
$$
\begin{align}
\sum_{k=0}^{n-1}e^{\frac{2\pi i}nk}
&=\sum_{k=0}^{n-1}\cos\left(\frac{2\pi}nk\right)+i\sum_{k=0}^{n-1}\sin\left(\frac{2\pi}nk\right)\tag{2}
\end{align}
$$
Substituting $n\mapsto2n+1$ and subtracting the $k=0$ term yields
$$
\sum_{k=1}^{2n}\cos\left(\frac{2\pi}{2n+1}k\right)=-1\tag{3}
$$
Substituting $k\mapsto2n+1-k$ shows that
$$
\sum_{k=1}^n\cos\left(\frac{2\pi}{2n+1}k\right)
=\sum_{k=n+1}^{2n}\cos\left(\frac{2\pi}{2n+1}k\right)\tag{4}
$$
Combine $(3)$ and $(4)$ to get the desired result.
Alternate Approach: Using the formula for the sum of a geometric series, we get
$$
\begin{align}
\sum_{k=1}^ne^{\frac{2\pi i}{2n+1}k}
&=\frac{e^{\frac{2\pi i}{2n+1}(n+1)}-e^{\frac{2\pi i}{2n+1}}}{e^{\frac{2\pi i}{2n+1}}-1}\\
&=\frac{e^{\pi i}-e^{\frac{\pi i}{2n+1}}}{e^{\frac{\pi i}{2n+1}}-e^{-\frac{\pi i}{2n+1}}}\\
&=\frac{-1-\cos\left(\frac\pi{2n+1}\right)-i\sin\left(\frac\pi{2n+1}\right)}{2i\sin\left(\frac\pi{2n+1}\right)}\\
&=\frac i2\cot\left(\frac\pi{4n+2}\right)\color{#C00000}{-\frac12}\tag{5}
\end{align}
$$
and using Euler's Formula
$$
\begin{align}
\sum_{k=1}^ne^{\frac{2\pi i}{2n+1}k}
&=\color{#C00000}{\sum_{k=1}^n\cos\left(\frac{2\pi}{2n+1}k\right)}+i\sum_{k=1}^n\sin\left(\frac{2\pi}{2n+1}k\right)\tag{6}
\end{align}
$$
The real parts of $(5)$ and $(6)$ give the desired result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1533626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$\lim_{x \to 0}\frac{\sin^2x-x^2}{x^2\sin^2x}$ $\lim_{x \to 0}\frac{\sin^2x-x^2}{x^2\sin^2x}$
I just can't do anything with this besides l'Hospital's rule (which doesn't seem to be a good idea). Can you help me, please?
| Since $x \approx \sin x$ when $x \rightarrow 0$, the denominator is of order $x^4$.
The numerator needs more careful analysis of $\sin x$. Using $\sin x = x - \frac{x^3}{6} + o(x^3)$, we get $\sin^2 x = x^2 - \frac{x^4}{3} + o(x^4)$, so the numerator is $\sin^2 x - x^2 \approx -\frac{x^4}{3}$.
Dividing and ignoring terms of lower order, we get the limit: $-\frac{1}{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1533745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve $y^2 = x^3 - 4$ where $x,y \in \mathbb{Z}$ Solve $y^2 = x^3 - 4$
So far i've tried to look at this:
$x^3 = (y-2i)(y+2i)$.
I think we need to prove that $y+-2i$ are coprime, so suppose not then:
$d = u+iv$ divides both numbers. So also d divides the sum and difference: $d|2y$ and $d|4i$.
If one applies the euclidean norm function we get that $u^2 + v^2|4y^2$ and $u^2 + v^2|16$.
And here i get stuck, any tips?
Kees
| Your computation are done in the ring of Gaussian integers, which is a PID. Let $p$ be a prime in this ring dividing both $y+2i$ and $y-2i$. Then $p$ divides $2y$ and $4i$, so it is $1+i$ (up to invertible elements), because this is the only prime dividing $2$.
From $(1+i)(h+ki)=y+2i$ we get $h-k=y$, $h+k=2$, so $h=(2+y)/2$, $k=(2-y)/2$ and $y$ must be even.
So the two elements are indeed coprime if $y$ is odd. In this case $y+2i$ must be a cube times an invertible element, so
$$
y+2i=u(a+bi)^3
$$
In the case $u=1$, we get $y=a(a^2-3b^2)$, $2=b(3a^2-b^2)$. Since $y$ is odd, $a$ must be odd and $b$ even. Thus $b=\pm2$ and $3a^2-4=\pm1$. The only case is $b=-2$, $a=\pm1$. So $y=\pm11$.
A solution has been found: $x=5,y=\pm11$.
In the case $u=i$, we get $y=-b(3a^2-b^2)$, $2=-a(a^2-3b^2)$. So $b$ must be odd and $a$ even. Therefore $a=\pm2$. If $a=2$ we get $4-3b^2=-1$ (no solution). If $a=-2$ we get $4-3b^2=1$, so $b=\pm1$. This corresponds to $y=\pm11$, the same as before.
The cases $u=-1$ and $u=-i$ are the same, because the minus sign can be merged with the cube.
Suppose $y$ is even. Then also $x$ must be even and so we can set $x=2X$ and $y=2Y$, obtaining
$$
2X^3=Y^2+1=(Y+i)(Y-i)
$$
Since $1+i$ is prime and divides the left-hand side, it must divide $Y+i$ or $Y-i$. From $(1+i)(h+ki)=Y+i$ we get $h-k=Y$ and $h+k=1$. Thus $h=(Y+1)/2$ and $Y$ must be odd. Similarly, if $1+i$ divides $Y-i$. If we set $Y=2Z+1$, we get
$$
\frac{2Z+1+i}{1+i}=Z+1-Zi,
\qquad
\frac{2Z+1-i}{1-i}=Z+1+Zi.
$$
and the equation can be written as
$$
X^3=(Z+1-Zi)(Z+1+Zi)
$$
A prime dividing both factors must divide $2(Z+1)$ and $2Zi$. If it divides $Z$, it cannot divide $Z+1$, so it divides $2$. So the only prime that can divide both numbers is $1+i$. However, from
$$
(1+i)(h+ki)=Z+1+Zi
$$
we get $2h=2Z+1$, a contradiction. So the two factors are coprime and therefore they are cubes times an invertible.
Case $Z+1+Zi=(a+bi)^3$. We get $Z+1=a(a^2-3b^2)$ and $Z=b(3a^2-b^2)$.
Therefore $a^3-3ab^2-3a^2b+b^3=1$, or $(a+b)(a^2-4ab+b^2)=1$.
From $a+b=1$ we get $(a+b)^2-6ab=1$, hence or $6ab=0$. Thus $a=0$ or $b=0$ and we get the solutions $a=0$, $b=1$, that is $Z=-1$; $a=1$, $b=0$, that is $Z=0$. This corresponds to $Y=-1$ and $Y=1$.
From $a+b=-1$ we get $-6ab=-2$, that has no solution.
Case $Z+1+Zi=i(a+bi)^3$. $Z+1=-b(3a^2-b^2)$ and $Z=a(a^2-3b^2)$. Here we get $-a^3-3a^2b+3ab^2+b^3=1$, so $(b-a)(a^2+4ab+b^2)=1$ The case $b-a=1$ forces $6ab=0$. The case $b-a=-1$ forces $6ab=2$. Thus we don't find new solutions.
Another solution has been found $x=2$, $y=\pm2$
No other solutions exist
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Is it possible to find solution of this system of equations? Following is augmented matrix which has been reduced to row echelon form by using row operations. So when I convert it to system of equations I would get 3 equations with 5 unknowns. Is it possible to find values of 5 unknowns in 3 equations? Is it true that at most one can solve equations with 3 unknowns in 3 equations?
$$
\left[
\begin{array}{rrrrr|r}
1 & 7& -2 & 0 & -8 & -3 \\
0 & 0 & 1 & 1 & 6 & 5 \\
0 & 0 & 0 & 1 & 3 & 9 \\
0 & 0 & 0 & 0 & 0 & 0 \\
\end{array}
\right]
$$
|
Is it possible to find values of 5 unknowns in 3 equations?
Yes, the solutions of your system form a 2 dimensional subspace of $F^5$, where $F$ is your field, e.g. $F = \mathbb{R}$.
$$
\left[
\begin{array}{rrrrr|r}
1 & 7 & -2 & 0 & -8 & -3 \\
0 & 0 & 1 & 1 & 6 & 5 \\
0 & 0 & 0 & 1 & 3 & 9 \\
0 & 0 & 0 & 0 & 0 & 0 \\
\end{array}
\right]
\to
\left[
\begin{array}{rrrrr|r}
1 & 7 & 0 & 2 & 4 & 7 \\
0 & 0 & 1 & 1 & 6 & 5 \\
0 & 0 & 0 & 1 & 3 & 9 \\
0 & 0 & 0 & 0 & 0 & 0 \\
\end{array}
\right]
\to
\left[
\begin{array}{rrrrr|r}
1 & 7 & 0 & 0 & -2 &-11 \\
0 & 0 & 1 & 0 & 3 & -4 \\
0 & 0 & 0 & 1 & 3 & 9 \\
0 & 0 & 0 & 0 & 0 & 0 \\
\end{array}
\right]
$$
E.g.
$$
x_5 = t \\
x_4 + 3t = 9 \iff x_4 = 9 - 3t \\
x_3 + 3t = -4 \iff x_3 = -4 - 3t \\
x_2 = s \\
x_1 + 7s - 2t = -11 \iff x_1 = -11 - 7s + 2t
$$
then
$$
x = \{ (-11-7s+2t, s, -4 - 3t, 9-3t, t) \mid s, t \in F \}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1534338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show that $x^2 \equiv 1 \pmod{n}$ can have arbitrarily many solutions? Equivalently, show that for every $m$, there exists some $n$ such that $x^2 \equiv 1 \pmod{n}$ has at least $m$ solutions.
Would appreciate any help or hints. I have a feeling this may be a pretty simple proof, but right now I am stuck. Thanks!
| Expanding on my comment:
Let $k$ be some positive integer such that $2^k \geq m$. Let $p_1, p_2, p_3, \cdots, p_k$ be $k$ distinct odd primes. (It's possible to choose $k$ distinct odd primes since there are infinitely many primes.)
Consider $n = p_1 p_2 p_3 \cdots p_k$, and look at the system of congruences
$$\begin{align*}
x \equiv \pm 1 & \pmod{p_1} \\
x \equiv \pm 1 & \pmod{p_2} \\
x \equiv \pm 1 & \pmod{p_3} \\
& \vdots \\
x \equiv \pm 1 & \pmod{p_k}
\end{align*}$$
where we choose either $+1$ or $-1$ for each prime. There are two choices for each prime, and so there are $2^k \geq m$ possible systems that can arise in this way.
by the Chinese Remainder Theorem, each one of these $2^k$ systems has some solution $x$ modulo $p_1 p_2 p_3 \cdots p_k = n$. No two systems give us the same value for $x$ modulo $n$, since if $x_1$ and $x_2$ arise from two different systems of congruences then there is some prime $p_i$ such that
$$ \begin{align*}
x_1 & \equiv \pm 1 \pmod {p_i} && \text{ but } & x_2 & \equiv \mp 1 \pmod {p_i}
\end{align*} $$
(i.e. $x_1$ and $x_2$ have different remainders modulo some prime factor of $n$)
This shows that $x_1$ and $x_2$ are not congruent modulo $n$, and so we in fact get $2^k$ distinct values of $x$ by looking at systems of congruences of the form given above.
Now for any such solution $x$, we have that
$$\begin{align*}
x^2 \equiv (\pm 1)^2 & \equiv 1 \pmod{p_1} \\
x^2 \equiv (\pm 1)^2 & \equiv 1 \pmod{p_2} \\
x^2 \equiv (\pm 1)^2 & \equiv 1 \pmod{p_3} \\
& \vdots \\
x^2 \equiv (\pm 1)^2 & \equiv 1 \pmod{p_k}
\end{align*}$$
and so we see (since the $p_i$ are pairwise relatively prime) that
$x^2 \equiv 1 \pmod n$.
Thus each of the $2^k \geq m$ possible values for $x$ are a solution to $x^2 \equiv 1 \pmod n$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluation of $ \int_{0}^{\infty}\frac{x\ln x}{(1+x^2)^2}dx$ $\bf{My\; Try:}$ Let $$\displaystyle I = \int_{0}^{\infty}\frac{x\ln x}{(1+x^2)^2}\,dx = \underbrace{\int_{0}^{1}\frac{x\ln x}{(1+x^2)^2}\,dx}_{I_{1}}+\underbrace{\int_{1}^{\infty}\frac{x\ln x}{(1+x^2)^2}\,dx}_{I_{2}}.$$
Now Here $$\displaystyle I_{2} = \int_{1}^{\infty}\frac{x\ln x}{(1+x^2)^2}\,dx.$$ Put $\displaystyle x=\frac{1}{t}.$ Then $\displaystyle dx = -\frac{1}{t^2}dt$ and changing limit
We get $$\displaystyle I_{2} = \int_{1}^{0}\frac{-t^3\cdot \ln t}{(1+t^2)^2}\cdot -\frac{1}{t^2}\,dt = \int_{1}^{0}\frac{t\ln t}{(1+t^2)^2}\,dt = -\int_{0}^{1}\frac{t\ln t}{(1+t^2)^2}\,dt$$
So we get $$\displaystyle I_{2} = -\int_{0}^{1}\frac{x\ln x}{(1+x^2)^2}dt = -I_{1}\Rightarrow I_{1}+I_{2} = 0$$
So we get $$\displaystyle I = \int_{0}^{\infty}\frac{x\ln x}{(1+x^2)^2}dx = I_{1}+I_{2} =0$$
My question is, can we solve it using another method? If so then please explain.
| Your solution is simple and straight to the point. Here is a longer way:
You could take the indefinite integral then apply limits to find the definite integral.
Let $v=\log x$, $u'=\frac{x}{(1+x^2)^2}$
So $v'=\frac{1}{x}$, $u=-\frac{1}{2(1+x^2)}$
$$\int\frac{{x\log x}}{(1+x^2)^2}dx=-\frac{\log x}{2(1+x^2)}+\int\frac{1}{2x(1+x^2)}dx$$
$$=-\frac{\log x}{2(1+x^2)}+\int\frac{1}{2x}-\frac{x}{2(1+x^2)}dx$$
$$=-\frac{\log x}{2(1+x^2)}+\frac{\log x}{2}-\frac{\log(1+x^2)}{4}$$
$$=-\frac{x^2\log x}{2(1+x^2)}-\frac{\log(1+x^2)}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1536946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
How to calculate this infinite sum? $$ \sum_ {n=0}^\infty \frac {1}{(4n+1)^2} $$
I am not sure how to calculate the value of this summation. My working so far is as follows:
Let $S=\sum_ {n=0}^\infty \frac {1}{(4n+1)^2}$.
$\Longrightarrow S=\frac{1}{1^2}+\frac{1}{5^2}+\frac{1}{9^2}+...$
$\Longrightarrow S=(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+...)-(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{6^2}+...) $
$\Longrightarrow S=\zeta(2)-[(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+\frac{1}{8^2}+\frac{1}{9^2}...)-(\frac{1}{5^2}+\frac{1}{9^2}+...)]$
$\Longrightarrow S=\zeta(2)-[(\zeta(2)-1)-(S-1)]$
$\Longrightarrow S=\zeta(2)-\zeta(2)+1+S-1$
$\Longrightarrow 0=0$
Does anyone have a better way of evaluating this that does not involve a cyclical answer as mine eventually does?
| This has no closed form, unless you want to use Catalan's constant. Then the answer is $\frac{K}{2}+\pi^2/16$. See here: http://mathworld.wolfram.com/CatalansConstant.html
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1537858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
The line $\frac{x+6}{5}=\frac{y+10}{3}=\frac{z+14}{8}$ is the hypotenuse of an isosceles right angled triangle whose opposite vertex is $(7,2,4)$ The line $\dfrac{x+6}{5}=\dfrac{y+10}{3}=\dfrac{z+14}{8}$ is the hypotenuse of an isosceles right angled triangle whose opposite vertex is $(7,2,4)$.Find the equation of the remaining sides.
My Attempt:
Let the right angled triangle be $ABC$,right angled at $B(7,2,4).$The remaining two sides $AB$ and $BC$ of the right angled triangle will pass through the vertex $B(7,2,4)$.
Let the equation of $AB$ be $\dfrac{x-7}{a_1}=\dfrac{y-2}{b_1}=\dfrac{z-4}{c_1}$
and the equation of $BC$ be $\dfrac{x-7}{a_2}=\dfrac{y-2}{b_2}=\dfrac{z-4}{c_2}$
As $AB$ and $BC$ are perpendicular to each other. So $a_1a_2+b_1b_2+c_1c_2=0.....(1)$
Also as the triangle is a isosceles right triangle, so angle between $AC$ and
$BC$ will be same as the angle between $CA$ and $BA$. Let it be $\theta$.
So $\cos\theta=\dfrac{5a_1+3b_1+8c_1}{\sqrt{5^2+3^2+8^2}\sqrt{a_1^2+b_1^2+c_1^2}}=\dfrac{5a_2+3b_2+8c_2}{\sqrt{5^2+3^2+8^2}\sqrt{a_2^2+b_2^2+c_2^2}}$
Squaring both sides we get
$\dfrac{(5a_1+3b_1+8c_1)^2}{a_1^2+b_1^2+c_1^2}=\dfrac{(5a_2+3b_2+8c_2)^2}{a_2^2+b_2^2+c_2^2}.............(2)$
But I am stuck here.I don't know how to find $a_1,b_1,c_1,a_2,b_2$ and $c_2$. Please help me. Thanks.
| Sides AB and AC make an angle of 45∘ with the given line.
Let B=(−6+5t,−10+3t,−14+8t)
AB→=(5t−13)i^+(3t−12)j^+(8t−18)k^
cos45∘=((5t−13)i^+(3t−12)j^+(8t−18)k^).(5i^+3j^+8k^)(5t−13)2+(3t−12)2+(8t−18)2−−−−−−−−−−−−−−−−−−−−−−−−−−−√52+32+82−−−−−−−−−−√
Squaring and simplifying t2–5t+6=0⟹t=2or3
Taking 2 for B and 3 for C.
AB→=−3i^−6j^−2k^
AC→=2i^−3j^+6k^
Equation of AB is x−73=y−26=z−42
Equation of AC is x−72=y−2−3=z−46
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1538543",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Prove that if $A\ge B$ then $\left[ {\begin{array}{*{20}{c}} A & B \\ B & A \\ \end{array}} \right]\ge0$. Let $A$ and $B$ be $n \times n$ matrices, i.e., $A, B \in M_n$. Also, $A \ge 0$, $B \ge 0$, and $A-B \ge 0$ which mean all these matrices are semi-positive-definite.
Why does
$\left[ {\begin{array}{*{20}{c}}
A & B \\
B & A \\
\end{array}} \right] \ge 0$?
| First, write
$$
\begin{pmatrix}
x' & y'
\end{pmatrix}
\begin{pmatrix}
A & B\\
B & A
\end{pmatrix}
\begin{pmatrix}
x \\ y
\end{pmatrix}
=x'Ax+y'Bx+x'By+y'Ay.
$$
Next, consider the RHS above
\begin{align*}
\text{RHS}&=x'(A-B)x+x'Bx+y'Bx+x'By+y'By+y'(A-B)y\\
&=x'(A-B)x+(x+y)'Bx+(x+y)'By+y'(A-B)y\\
&=x'(A-B)x+(x+y)'B(x+y)+y'(A-B)y
\end{align*}
which is nonnegative due to the positive semidefiniteness of $A-B$ and $B$. The claim follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1539605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
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