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solve$\frac{xdx+ydy}{xdy-ydx}=\sqrt{\frac{a^2-x^2-y^2}{x^2+y^2}}$ solve the differential equation.
$$\frac{xdx+ydy}{xdy-ydx}=\sqrt{\frac{a^2-x^2-y^2}{x^2+y^2}}$$
The question is from IIT entrance exam paper. I have tried substituting $x^2=t\ and \ y^2=u$ but was not a worth try.
Thanks in advance.
|
This equation could have been solved easily in polar coordinates.
\begin{align}
x &= z \cos{s} \\
y &= z \sin{s}
\end{align}
Then
\begin{align}
dx = \cos(s) \, dz - z\,\sin(s)\, ds \\
dy = \sin(s) \, dz + z\,\cos(s)\, ds
\end{align}
as well as
$$x^2+y^2 = z^2\big(\cos^2(s) + \sin^2(s)\big) = z^2$$ which means that
$$z = \sqrt{x^2 + y^2}$$
Now
\begin{align}
y\,dx = z \sin(s) \cos(s) \, dz - z^2\,\sin^2(s)\, ds \\
x\,dy = z \cos(s) \sin(s) \, dz + z^2\,\cos^2(s)\, ds
\end{align}
which leads to the difference
\begin{align}
x\,dy - y\,dx &= z \cos(s) \sin(s) \, dz + z^2\,\cos^2(s)\, ds\\
& - z \sin(s) \cosh(s) \, dz + z^2\,\sinh^2(s)\, ds \\
&= z^2\,\cos^2(s)\, ds + z^2\,\sin^2(s)\, ds \\
&= z^2\big(\cos^2(s) + \sin^2(s)\big) \, ds\\
&= z^2\, ds
\end{align}
Moreover,
\begin{align}
x\,dx + y\,dy &= \frac{1}{2}d \left(x^2+y^2\right)\\
&= \frac{1}{2} d(z^2)\\
&= z\,dz
\end{align}
Consequently
\begin{align}\frac{x\,dx + y\,dy}{ x\,dy - y\,dx } &= \frac{z\, dz}{z^2 \, ds}\\ &= \frac{1}{z} \frac{dz}{ds}\end{align}
and finaly the equations becomes
\begin{align}
\frac{1}{z} \frac{dz}{ds} = \frac{x\,dx + y\,dy}{ x\,dy - y\,dx } = \sqrt{\frac{a^2 - x^2 - y^2}{x^2+y^2}} = \sqrt{\frac{a^2 - z^2}{z^2}} = \frac{\sqrt{a^2-z^2}}{z}
\end{align}
multiply both sides by $z$ and obtain the simple differential equation
$$ \frac{dz}{ds} = \sqrt{a^2-z^2}.$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/1015021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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|
Convergence of $\sum_{n=1}^{\infty}\frac{1}{n}\Big(\sqrt[3]{(n+1)^{2}} - \sqrt[3]{n^{2}}\Big)$ $\displaystyle\sum_{n=1}^{\infty}\frac{\sqrt[3]{(n+1)^{2}} - \sqrt[3]{n^{2}}}{n}$
Converging or Diverging? I guess I have to lower the fraction so that the roots will get away and I will have $\frac{1} {n}$ that diverges. But I have no idea how to do that.
Any ideas?
|
$$
0<\frac{\sqrt[3]{(n+1)^{2}} - \sqrt[3]{n^{2}}}{n}=
\frac{\sqrt[3]{(n+1)^{2}} - \sqrt[3]{n^{2}}}{n}\cdot
\frac{\sqrt[3]{(n+1)^{4}} +\sqrt[3]{(n+1)^{2}}\sqrt[3]{n^{2}}+ \sqrt[3]{n^{4}}}
{\sqrt[3]{(n+1)^{4}} +\sqrt[3]{(n+1)^{2}}\sqrt[3]{n^{2}}+ \sqrt[3]{n^{4}}} \\
=\frac{(n+1)^2-n^2}
{
n(\sqrt[3]{(n+1)^{4}} +\sqrt[3]{(n+1)^{2}}\sqrt[3]{n^{2}}+\sqrt[3]{n^4})
}\le \frac{2n+1}{n\cdot 3n^{4/3}}\le\frac{3n}{3n^{7/3}}=\frac{3}{n^{4/3}}.
$$
Hence, due to the Comparison Test,
the series converge, as $\sum_{n=1}^\infty \frac{1}{n^a}$, converges whenever $a>1$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1016466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Inequality with five variables Let $a$, $b$, $c$, $d$ and $e$ be positive numbers. Prove that:
$$\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+d}+\frac{d}{d+e}+\frac{e}{e+a}\geq\frac{a+b+c+d+e}{a+b+c+d+e-3\sqrt[5]{abcde}}$$
Easy to show that $$\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\geq\frac{a+b+c}{a+b+c-\sqrt[3]{abc}}$$ is true
and for even $n$ and positives $a_i$ the following inequality is true.
$$\frac{a_1}{a_1+a_2}+\frac{a_2}{a_2+a_3}+...+\frac{a_n}{a_n+a_1}\geq\frac{a_1+a_2+...+a_n}{a_1+a_2+...+a_n-(n-2)\sqrt[n]{a_1a_2...a_n}}$$
|
A proof for $n=3$.
We'll prove that $\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\geq\frac{a+b+c}{a+b+c-\sqrt[3]{abc}}$ for all positives $a$, $b$ and $c$.
Indeed, let $ab+ac+bc\geq(a+b+c)\sqrt[3]{abc}$.
Hence, by C-S $\sum\limits_{cyc}\frac{a}{a+b}\geq\frac{(a+b+c)^2}{\sum\limits_{cyc}(a^2+ab)}=\frac{1}{1-\frac{ab+ac+bc}{(a+b+c)^2}}\geq\frac{a+b+c}{a+b+c-\sqrt[3]{abc}}$.
Let $ab+ac+bc\leq(a+b+c)\sqrt[3]{abc}$.
Hence, by C-S $\sum\limits_{cyc}\frac{a}{a+b}\geq\frac{(ab+ac+bc)^2}{\sum\limits_{cyc}(a^2c^2+a^2bc)}=\frac{1}{1-\frac{abc(a+b+c)}{(ab+ac+bc)^2}}\geq\frac{1}{1-\frac{\sqrt[3]{abc}}{a+b+c}}=\frac{a+b+c}{a+b+c-\sqrt[3]{abc}}$.
Done!
|
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"url": "https://math.stackexchange.com/questions/1017110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "25",
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|
Basis vector find Let $\alpha_1=[ 2,1,3,0] $
$\alpha_2=[ 1,1,1,-1] $, $\alpha_3=[ 2,-1,5,4] $, $\alpha_4=[ 1,2,0,-3] $, $\alpha_5=[ 3,1,6,1] $
be vectors from $\mathbb{R}^4$ . From vectors system ($\alpha_1,\alpha_2, \alpha_3, \alpha_4, \alpha_5 $) choose basis of vector space $V=lin(\alpha_1,\alpha_2, \alpha_3, \alpha_4, \alpha_5)\subset\mathbb{R}^4$ spanned by those vectors.
I need help with creating a proper matrix for this question.
|
Consider the matrix
\begin{bmatrix}
2 & 1 & 2 & 1 & 3\\
1 & 1 & −1 & 2 & 1\\
3 & 1 & 5 & 0 & 6\\
0 & −1 & 4 & −3 & 1
\end{bmatrix}
and perform Gaussian elimination on it:
\begin{align}
\begin{bmatrix}
2 & 1 & 2 & 1 & 3\\
1 & 1 & −1 & 2 & 1\\
3 & 1 & 5 & 0 & 6\\
0 & −1 & 4 & −3 & 1
\end{bmatrix}
&\to
\begin{bmatrix}
1 & 1/2 & 1 & 1/2 & 3/2\\
1 & 1 & −1 & 2 & 1\\
3 & 1 & 5 & 0 & 6\\
0 & −1 & 4 & −3 & 1
\end{bmatrix}
\\&\to
\begin{bmatrix}
1 & 1/2 & 1 & 1/2 & 3/2\\
0 & 1/2 & −2 & 3/2 & -1/2\\
0 & -1/2 & 2 & -3/2 & 3/2\\
0 & −1 & 4 & −3 & 1
\end{bmatrix}
\\&\to
\begin{bmatrix}
1 & 1/2 & 1 & 1/2 & 3/2\\
0 & 1 & −4 & 3 & -1\\
0 & 0 & 0 & 0 & 1\\
0 & 0 & 0 & 0 & 0
\end{bmatrix}
\end{align}
The columns where a pivot is found are linearly independent and the others can be written as linear combination of them. So a basis is
$$
\{\alpha_1,\alpha_2,\alpha_5\}
$$
It's not the only solution, of course, but you can see that $\alpha_5$ will be in any basis extracted from that set, because it is not a linear combination of the other four vectors. If you go on with backwards elimination,
\begin{align}
\begin{bmatrix}
1 & 1/2 & 1 & 1/2 & 3/2\\
0 & 1 & −4 & 3 & -1\\
0 & 0 & 0 & 0 & 1\\
0 & 0 & 0 & 0 & 0
\end{bmatrix}
&\to
\begin{bmatrix}
1 & 1/2 & 1 & 1/2 & 0\\
0 & 1 & −4 & 3 & 0\\
0 & 0 & 0 & 0 & 1\\
0 & 0 & 0 & 0 & 0
\end{bmatrix}
\\&\to
\begin{bmatrix}
1 & 0 & 3 & -1 & 0\\
0 & 1 & −4 & 3 & 0\\
0 & 0 & 0 & 0 & 1\\
0 & 0 & 0 & 0 & 0
\end{bmatrix}
\end{align}
you see that
\begin{align}
\alpha_3&=3\alpha_1-4\alpha_2\\
\alpha_4&=-\alpha_1+3\alpha_2
\end{align}
because elementary row operation don't change linear relations between columns.
|
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"url": "https://math.stackexchange.com/questions/1019078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Evaluation of the integral $\int \sqrt{t^4-t^2 + 1}\,dt$ My friend took his Calculus $2/3$ test yesterday.
One of the questions he had trouble with was this integral:
$$\int \sqrt{t^4-t^2 + 1}dt$$
My attempt
It seems rather clear that the only approach was trigonometric substitution.
First, completing the square:
$$\int \sqrt{t^4-t^2 + 1}dt = \int\sqrt{t^4-t^2 + \frac{1}{4} + \frac{3}{4}}dt = \int \sqrt{\left(t^2 - \frac{1}{2}\right) + \frac{3}{4}}dt$$
Next, I let $$\sec \theta = \frac{\sqrt{\left(t^2 - \frac{1}{2}\right) + \frac{3}{4}}}{\frac{\sqrt{3}}{2}}$$
$$\frac{\sqrt{3}}{2}\sec \theta = \sqrt{\left(t^2 - \frac{1}{2}\right) + \frac{3}{4}}$$
$$\tan \theta = \frac{t^2 - \frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{2t^2 - 1}{\sqrt{3}}$$
$$\sec^2 \theta d\theta = \frac{4}{\sqrt{3}}tdt,\ dt = \frac{\sqrt{3}\sec^2 \theta}{4t} d\theta$$
$$t = \sqrt{\frac{1}{2}\left(\sqrt{3}\tan \theta + 1\right)}$$
Substituting this all in:
$$\int \left(\frac{\sqrt{3}}{2} \sec \theta\right) \frac{\sqrt{3}\sec^2 \theta}{4\sqrt{\frac{1}{2}\left(\sqrt{3}\tan \theta + 1\right)}} d\theta$$
How would I approach this from here?
I'm thinking of using u-substitution but I'm sure that it would bring me back to where I started, meaning I would have to use trig. substitution again.
|
The indefinite integral cannot be expressed in terms of elementary functions. Evaluating it requires knowledge of elliptic integrals. However, it is interesting to note that its definite counterpart, when evaluated over the entire real line, and subtracted from its quadratic or parabolic asymptote, yields
$$\int_{-\infty}^\infty\bigg[\sqrt{t^4-t^2+1}-\bigg(t^2-\dfrac12\bigg)\bigg]dt~=~\dfrac23K\bigg(\dfrac34\bigg)+\dfrac43E\bigg(\dfrac34\bigg),$$
which is nothing else than the volume of the oloid, whose decimal expansion can be found on OEIS
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
summation of $\sum^n_{k=0} (n-k)^2$ I'm trying to find the recurrence of
$$ T(n) = T (n-1) + n^2$$
After following the steps,
$$T (n) = T (n-1) + n^2 = T (n-2) + (n-1)^2 + n^2 $$
$$T (n) = T (n-2) + (n-1)^2 + n^2 = T(n-3) + (n-3)^2 + (n-1)^2 + n^2 $$
$$T (n) = T (n-3) + (n-3)^2 + (n-1)^2 + n^2 =T (n-4) + (n-4)^2 + (n-3)^2 + (n-1)^2 + n^2 $$
i generalize recurrence relation at the kth step of the recursion, which is
$$ T(n) = T (n-k) + \sum^n_{k=0} (n-k)^2$$
Just wondering,
What is the summation of
$$\sum^n_{k=0} (n-k)^2$$
is it the same as $$ n(n+1)(2n+1)/6$$
?
|
Without expanding, a simple substitution will help show that the result is true.
Put $r=n-k$, in which case when $k=0,n$ corresponds to $r=n,0$.
Hence
$$\sum_{k=0}^{n}(n-k)^2=\sum_{r=0}^n r^2=\frac16n(n+1)(2n+1)$$
NB: In fact, the original recurrence relationship telescopes quite readily as follows, without having to resort to iterative expansion or summing $(n-k)^2$:
$$\begin{align}
T(n)-T(n-1)&=n^2\\
T(n-1)-T(n-2)&=(n-1)^2\\
&\vdots\\
T(1)-T(0)&=1^2\end{align}$$
Summing by telescoping and taking $T(0)=0$ gives
$$T(n)=\sum_{r=1}^n r^2=\frac16n(n+1)(2n+1)$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/1020993",
"timestamp": "2023-03-29T00:00:00",
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|
If $P(x)$ is a poly. of least degree and local Max. at $x=1$ and Local Min. at $x=3,$ Then $P'(0)$ is If $P(x)$ is a polynomial of least degree which has a local Maxima at $x=1$ and Local Minima at
$x=3.$ If $P(1)=6$ and $P(3)=2$. Then $P'(0)=$
$\bf{My\; Try::}$ Given function has one Maxima and one Minima So $P(x)$ must have least
degree $3$ polynomial. So Let $P(x)=Ax^3+Bx^2+Cx+D$ and $P'(x)=3Ax^2+2Bx+C.$
Now Given $P(1)=6\Rightarrow A+B+C+D = 6....................(1)$
and Given $P(3)=2\Rightarrow 27A+9B+3C+D=2..............(2)$
and Given $P'(1)=0\Rightarrow 3A+2B+C = 0.........................(3)$
and Given $P'(3)=0\Rightarrow 9A+6B+C = 0.........................(4)$
Now Subtract $(4)-(3)$, We Get $6A+4B=0\Rightarrow 3A+2B=0$
Similarly Sub $(2)-(1)\;,$ We Get $26A+8B+2C=-4\Rightarrow 13A+4B+C=-2$
Is there is any other method my which we can solve the above question in less complex way.
plz explain me, Thanks
|
Continuing from what you have got,
Subtracting $13A + 4B + C = -2$ from $3A + 2B + C = 0$ yields $10A + 2B = -2$.
Now we have $3A + 2B = 0$ and $10A + 2B = -2$.
Hence $A = -2/7$ and $B$, $C$, $D$ can be calculated by putting back into $3A + 2B = 0$, $3A + 2B + C = 0$ and $A + B + C + D = 6$ in order.
Then $P'(0)$ can be easily obtained.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1021189",
"timestamp": "2023-03-29T00:00:00",
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|
How to evaluate $\sum_{\gcd (p,q)=1} \frac{1}{p^2q^2}$? How do find the following sum
$$
\sum_{\gcd (p,q)=1} \frac{1}{p^2q^2}
$$
|
Let's look at the more general case,
$$\sum_{\gcd (p,q) = d} \frac{1}{p^2q^2}.$$
$\gcd (p,q) = d$ means we have $p = d\cdot a$ and $q = d\cdot b$ with $\gcd(a,b) = 1$, so we obtain
$$\sum_{\gcd(p,q) = d} \frac{1}{p^2q^2} = \sum_{\gcd(a,b) = 1} \frac{1}{(da)^2(db)^2} = \frac{1}{d^4}\sum_{\gcd(a,b) = 1} \frac{1}{a^2b^2}.$$
Now, any pair of positive integers occurs in exactly one of the sets $A_d = \{ (k,m) : \gcd(k,m) = d\}$, hence
\begin{align}
\left(\sum_{d=1}^\infty \frac{1}{d^4}\right)\left(\sum_{\gcd(p,q) = 1} \frac{1}{p^2q^2}\right)
&= \sum_{d=1}^\infty \sum_{\gcd(a,b) = d} \frac{1}{a^2b^2}\\
&= \sum_{a,b \in \mathbb{N}\setminus\{0\}} \frac{1}{a^2b^2}\\
&= \left(\sum_{n=1}^\infty \frac{1}{n^2}\right)^2.
\end{align}
Thus we obtain
$$\sum_{\gcd(p,q) = 1}\frac{1}{p^2q^2} = \frac{\left(\sum_{n=1}^\infty \frac{1}{n^2}\right)^2}{\sum_{d=1}^\infty \frac{1}{d^4}} = \frac{\zeta(2)^2}{\zeta(4)} = \frac{\pi^4/36}{\pi^4/90} = \frac{90}{36} = \frac{5}{2}.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Squeeze theorem: sequence of sums So I've been having some problems with finding the limit of this sequence:
$$\lim_{n \rightarrow \infty}\left(\frac{4}{ \sqrt[5]{1^{10}+1 }} + \frac{8}{\sqrt[5]{2^{10}+2}} + ... + \frac{4n}{\sqrt[5]{n^{10}+n}}\right)$$
I've tried to find it using the squeeze theorem, here's what I've tried:
$$ \frac{4 + 8 + ... + 4n}{\sqrt[5]{n^{10}+n}} \leq \frac{4}{ \sqrt[5]{1^{10}+1 }} + \frac{8}{\sqrt[5]{2^{10}+2}} + ... + \frac{4n}{\sqrt[5]{n^{10}+n}} \leq \frac{4 + 8 + ... + 4n}{\sqrt[5]{n^{10}}} $$
and:
$$\lim_{n \rightarrow \infty} \frac{4 + 8 + ... + 4n}{\sqrt[5]{n^{10}+n}} = \lim_{n \rightarrow \infty} \frac{\frac{4+4n}{2}\cdot n }{\sqrt[5]{n^{10}+n}} = \lim_{n \rightarrow \infty} \frac{2n^{2}+2}{\sqrt[5]{n^{10}+n}} = 2 $$
$$\lim_{n \rightarrow \infty} \frac{4 + 8 + ... + 4n}{\sqrt[5]{n^{10}}} = \lim_{n \rightarrow \infty} \frac{\frac{4+4n}{2}\cdot n }{\sqrt[5]{n^{10}}} = \lim_{n \rightarrow \infty} \frac{2n^{2}+2}{\sqrt[5]{n^{10}}} = 2 $$
Therefore the limit of my initial sequence should also be 2. But I find it highly unlikely, mostly because:
$$ \sum_{n=1}^{\infty} \frac{4n}{\sqrt[5]{n^{10}+n}} \text{ diverges}$$
and:
$$ \frac{4}{\sqrt[5]{1^{10}+1}} \approx 3.482202253184496556545...$$
So I must have chosen the wrong sequences to bound my original one. Where did I go wrong?
|
Consider $$s_n=\frac{4 n}{\sqrt[5]{n^{10}+n}}=\frac{4 n}{n^2\sqrt[5]{1+\frac{1}{n^9}}}=\frac{4}{n} \frac{1}{\sqrt[5]{1+\frac{1}{n^9}}}$$ So, for large values of $n$, it write $$s_n=\frac{4}{n}-\frac{4}{5 n^{10}}+\frac{12}{25
n^{19}}+O\left(\left(\frac{1}{n}\right)^{21}\right)$$ and, so, the sum $\sum_{n=1}^{\infty}s_n$ looks very much to the harmonic number which goes to $\infty$.
|
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|
Strategy for solving $7\vert2^{n+2}+3^{2n+1}$ by induction. So I have to show the following to be true using induction
$7\mid 2^{n+2}+3^{2n+1}$
This is easily checked with the case $n=0$ because $7 \mid 7$, but I assuming this holds for$n=k :$
$$7\mid 2^{k+2}+3^{2k+1}$$
I fail to see how I can use this for $n=(k+1):$
$$\begin{align*}\begin{split}
&7 \mid \left(2^{k+3}+3^{2k+3}\right) &\Longleftrightarrow\\
&7 \mid \left(2\times2^{k+2}+3^2\times3^{2k+1}\right)
\end{split}\end{align*}
$$
I'm assuming my strategy is flawed, but it's a while since I've done these and I can't remember what should be done.
|
You checked it for $n=0$, so now you have to prove implication:
$$ \left(\forall n \in \mathbb{N}\right)\left(\underbrace{7 \mid 2^{n+2} + 3^{2n+1}}_{\mathrm{Assumption}} \Longrightarrow
\underbrace{7 \mid 2^{(n+1)+2} + 3^{2(n+1)+1}}_{\mathrm{Thesis}}\right)$$
Simple thesis transformation:
$$\begin{align*}\begin{split}
2^{(n+1)+2} + 3^{2(n+1)+1} &= 2^1 \cdot 2^{n+2} + 3^2 \cdot 3^{2n+1}\\
&= 2 \cdot 2^{n+2} + 9 \cdot 3^{2n+1}\\
&= 2 \cdot 2^{n+2} + (2+7) \cdot 3^{2n+1}\\
&= \left(2 \cdot 2^{n+2} + 2\cdot 3^{2n+1}\right) +
7 \cdot 3^{2n+1}\\
&=2\cdot\left(2^{n+2} + 3^{2n+1} \right) + 7 \cdot
3^{2n+1}
\end{split}\end{align*}$$
From assumption $(\exists k \in \mathbb{Z})(2^{n+2} + 3^{2n+1} =7k)$, so
$$\begin{align*}\begin{split}
2^{(n+1)+2} + 3^{2(n+1)+1} &= 2\cdot\left(2^{n+2} + 3^{2n+1} \right) + 7 \cdot
3^{2n+1}\\
&= 2 \cdot 7k + 7 \cdot 3^{2n+1} \\
&= 7 \cdot \left(2\cdot k + 3^{2n+1}\right)
\end{split}\end{align*}$$
$$7 \mid \left(7 \cdot \left(2\cdot k + 3^{2n+1}\right) \right)
\Longleftrightarrow
7 \mid \left(2^{(n+1)+2} + 3^{2(n+1)+1}\right)$$
Which proves the inductive thesis, and we obtain $\left(\forall n \in \mathbb{N}\right)\left(7 \mid 2^{n+2} + 3^{2n+1} \right)$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove $\lim_{x\to-2}\frac{x+8}{x+3}=6 $ using $\epsilon-\delta$
Prove:
$$\lim_{x\to-2}\frac{x+8}{x+3}=6 $$
I started with:
$ |\frac{x+8}{x+3} - 6| < \epsilon => |\frac{x+8-6x-18}{x+3} | < \epsilon => |\frac{-5x-10}{x+3} | < \epsilon =>| \frac{-5(x+2)}{x+3} | < \epsilon $
From the definition of limit we know that:
$ |x+2| <\delta $
Just don't know how to continue now... any suggestions?
|
Hint:
Take $\delta=min(\frac{1}{2},\frac{\epsilon}{10})>0$
Then $\forall x, |x+2| <\delta $, $|x+3|>\frac{1}{2}$, hence $$| \frac{-5(x+2)}{x+3} |\le 10|x+2| < \epsilon $$
The key is to select $\delta$ such that $x$ is closed to $-2$ compared with $-3$.
|
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|
The inequality $ abc\geq(a-b+c)(a+b-c)(b+c-a)$ holds for $a,b,c\geq 0$
What is the proof that:
$$\forall \ a,b,c\geq 0:\quad a\ b\ c\geq(a-b+c)(a+b-c)(b+c-a)$$
I tried :
we can write that expression $(a-b+c)(a+b-c)(b+c-a)$ as $$-a^3+a^2 b+a^2 c+a b^2-2 a b c+a c^2-b^3+b^2 c+b c^2-c^3$$
then
$$a\ b\ c\geq(a-b+c)(a+b-c)(b+c-a)$$
$$\Longleftrightarrow $$
$$a b c \geq -a^3+a^2 b+a^2 c+a b^2-2 a b c+a c^2-b^3+b^2 c+b c^2-c^3$$
Now consider $f(a)=abc-(-a^3+a^2 b+a^2 c+a b^2-2 a b c+a c^2-b^3+b^2 c+b c^2-c^3)$ for $b,c$ fixed
if i follow the metode of Quang Hoang
let $a+b-c=2x,b+c-a=2y,c+a-b=2z$, then the inequality becomes
$$(y+z)(z+x)(x+y)\ge 8xyz.\tag{1}$$
for $x,y,z \geq 0$
we use AM-GM inequality
$$(y+z)(z+x)(x+y)\ge 2(yz)^{1/2}(2zx)^{1/2}(2xy)^{1/2}.\tag{1}$$
but other case of $x,y,z$ is not clear.
*
*i will be appreciated if someone could explain it with more detail
|
Write $a+b-c=2x,b+c-a=2y,c+a-b=2z$, then the inequality becomes
$$(y+z)(z+x)(x+y)\ge 8xyz.\tag{1}$$
Note that at most one of $x,y,z$ can be negative (consider the largest of $a,b,c$).
Edit: Without loss of generality, assume that $a=\max(a,b,c)$, then $$2x=(a-c)+b\ge b\ge 0, 2z=(a-b)+c\ge c\ge 0.$$
So there are two cases
*
*If $y\le0$: (1) is clear since the LHS is $\ge 0$, the RHS is $\le 0$.
*If $y\ge 0$, then all $x,y,z$ are $\ge 0$. Using AM-GM:
$$x+y\ge 2\sqrt{xy},y+z\ge 2\sqrt{yz},z+x\ge 2\sqrt{zx}.$$
Multiplying those gives (1).
|
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|
How to evaluate $\lim_{n\to\infty}\sqrt[n]{\frac{1\cdot3\cdot5\cdot\ldots\cdot(2n-1)}{2\cdot4\cdot6\cdot\ldots\cdot2n}}$ Im tempted to say that the limit of this sequence is 1 because infinite root of infinite number is close to 1 but maybe Im mising here something? What will be inside the root?
This is the sequence:
$$\lim_{n\to\infty}\sqrt[n]{\frac{1\cdot3\cdot5\cdot\ldots\cdot(2n-1)}{2\cdot4\cdot6\cdot\ldots\cdot2n}}$$
|
Using the following inequality
\begin{equation}
\frac{n}{\sum_{k=1}^{n} \frac{1}{1-\frac{1}{2k}} } \leq \sqrt[n]{(1-\frac{1}{2})(1-\frac{1}{4})...(1-\frac{1}{2n})} \leq \frac{(1-\frac{1}{2})+(1-\frac{1}{4})+...,(1-\frac{1}{2n})}{n} = 1-\frac{1}{2n}\sum_{k=1}^{n}\frac{1}{k}
\end{equation}
and another results
\begin{equation}
\sum_{k=1}^{n}\frac{1}{k} = \ln(n) + \gamma ,n \rightarrow \infty,
\end{equation}
where $\gamma $ is Euler constant.
and
\begin{equation}
\sum_{k=1}^{n} \frac{1}{1-\frac{1}{2k}}=\sum_{k=1}^{n}\frac{2k-1 +1}{2k-1} = n + \sum_{k=1}^{n}\frac{1}{2k-1}\leq n + \sum_{k=1}^{n}\frac{1}{k}
\end{equation}
Hence we have
\begin{equation}
\frac{n}{n+\ln(n) + \gamma} \leq \sqrt[n]{(1-\frac{1}{2})(1-\frac{1}{4})...(1-\frac{1}{2n})} \leq 1 - \frac{\ln(n) + \gamma}{2n}
\end{equation}
|
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|
Use Mean Value Theorem to establish that $7 \frac{1}{4} < \sqrt{53} < 7 \frac{2}{7}$ I am working on the following question from Lay's Introduction to Analysis with Proof:
Use the Mean Value Theorem to establish that $7 \frac{1}{4} < \sqrt{53} < 7 \frac{2}{7}$
My work so far:
Define $f(t) = \sqrt t$ which we will work with on the interval $[49, 53]$ as $\sqrt{53} \sim \sqrt{49} = 7$.
By MVT,
$$(53-49)\frac{1}{2\sqrt x} = \sqrt{53} - \sqrt{49}$$
Now, $x \in [49, 53]$ so that $49 < x < 53 \Rightarrow 7 < \sqrt x < \sqrt{53} < 8 \Rightarrow \frac{1}{4} < \sqrt{53} - 7 < \frac{2}{7}$
So I have obtained the bound
$$ \frac{1}{4} + 7 < \sqrt{53} < \frac{2}{7} + 7$$
but don't know how to proceed as to show that $7 \frac{1}{4} < \sqrt{53} < 7 \frac{2}{7}$.
|
Notice that
$$7\frac{1}{4}=7+\frac{1}{4}$$
and
$$7\frac{2}{7}=7+\frac{2}{7}$$
|
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|
Compute: $\lim_{n \to \infty } \left ( 1-\frac{2}{3} \right ) ^{\frac{3}{n}}\cdots \left ( 1-\frac{2}{n+2} \right ) ^{\frac{n+2}{n}}$ Help me please to compute the limit of:
$
\lim_{n \to \infty } \left ( 1-\frac{2}{3} \right ) ^{\frac{3}{n}}\cdot \left ( 1-\frac{2}{4} \right ) ^{\frac{4}{n}}\cdot \left ( 1-\frac{2}{5} \right ) ^{\frac{5}{n}}\cdots \left ( 1-\frac{2}{n+2} \right ) ^{\frac{n+2}{n}}
$
What I did:
$
L=\frac{a_{n+1}}{a_n} = \frac{ \left ( 1-\frac{2}{n+3} \right ) ^{\frac{n+3}{n+1}}}{ \left ( 1-\frac{2}{n+2} \right ) ^{\frac{n+2}{n}}}=\frac{2/e}{2/e}=1
$
But it's not. Since $L=1$, I need use something else...
|
$(1-2/k)^k$ approaches $e^{-2}$, so most of the factors are near $e^{-2}$.
Then you take the $n$th root of them all, and there are $n$ factors, so the answer is $e^{-2}$
|
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|
Find $\log_c{x}$ if $\log_a{x} = p$, $\log_b{x} = q$, and $\log_{abc} {x} = r$.
Given that $\log_a{x} = p$, $\log_b{x} = q$, and $\log_{abc} {x} = r$, find the value of $\log_c{x}$.
|
Using the change of base and product rule for logs, we have:
$$
p = \frac{\log x}{\log a}
\qquad\text{and}\qquad
q = \frac{\log x}{\log b}
\qquad\text{and}\qquad
r = \frac{\log x}{\log abc} = \frac{\log x}{\log a + \log b + \log c}
$$
Taking reciprocals of each equation, we can combine them to obtain:
\begin{align*}
\frac{1}{r} - \frac{1}{q} - \frac{1}{p}
= \frac{\log a + \log b + \log c}{\log x} - \frac{\log b}{\log x} - \frac{\log a}{\log x}
= \frac{\log c}{\log x}
\end{align*}
Taking reciprocals again, we conclude that:
$$
\log_c x = \dfrac{\log x}{\log c} = \boxed{ \dfrac{1}{\frac{1}{r} - \frac{1}{q} - \frac{1}{p}}}
$$
|
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|
Show that $2(a^3+b^3+c^3)>a^2(b+c)+b^2(c+a)+c^2(a+b)>6abc$ If $a,b,c$ are positive real numbers, not all equal, then prove that
$$2(a^3+b^3+c^3)>a^2(b+c)+b^2(c+a)+c^2(a+b)>6abc$$
How can I show this?
|
*
*By my answer here and with the condition given,
$$2\left(a^3+b^3+c^3\right)>2\left(a^2b+b^2c+c^2a\right)$$
$\left(a^2b+b^2c+c^2a\right)>\left(ab^2+bc^2+ca^2\right)\\\impliedby ab(a-b)+bc(b-c)-ca(a-b)-ca(b-c)>0\\ \impliedby a(b-c)(a-b)-c(a-b)(b-c)>0 \\\impliedby (a-c)(a-b)(b-c)>0$
$$\color{blue}{\boxed{\color{blue}{\therefore2\left(a^3+b^3+c^3\right)>2\left(a^2b+b^2c+c^2a\right)>ab(a+b)+bc(b+c)+ca(c+a)}}}$$
*$ab^2+bc^2+ca^2>3abc\\\impliedby ab\left(b-c\right)+bc\left(c-a\right)+ca\left(a-b\right)>0\\\impliedby ab\left(b-c\right)-bc\left(a-b\right)-bc\left(b-c\right)+ca\left(a-b\right)>0\\\impliedby b(a-c)(b-c)+c(a-b)^2>0$
$$\color{blue}{\boxed{\color{blue}{\therefore ab(a+b)+bc(b+c)+ca(c+a)>6abc}}}$$
|
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|
Degree-4 Polynomial Solve the equation $x^4 - 14x^3 + 50x^2 -14x + 1 = 0$. I am not sure about how to best proceed, and would like a solution that does not involved the generalised quartic formula.
|
A more detailed solution:
If we divide the equation by $x^2$:
$$\frac{x^4}{x^2} - \frac{14x^3}{x^2} + \frac{50x^2}{x^2} - \frac{14x}{x^2} + \frac{1}{x^2} = x^2 - 14x + 50 - \frac{14}{x} + \frac{1}{x^2}$$
Then, combining like terms, we notice that:
$$x^2 + \frac{1}{x^2} - 14\left(x+\frac{1}{x}\right) + 50$$
If we let $y = x+\frac{1}{x}$
Note that:
$$\left(x+\frac{1}{x}\right)^2 = x^2 + \frac{1}{x^2} - 2$$
Therefore,
$$x^2 + \frac{1}{x^2} - 14\left(x+\frac{1}{x}\right) + 50 = y^2 -2 -14y + 50 =0$$
$$y^2 - 14y +48 =0$$
$$(y-6)(y-8) = 0$$
Therefore, $x + \frac{1}{x} = 6$ and $x + \frac{1}{x} = 8$
Can you take it from here?
|
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|
Factorising quadratics - coefficient of $x^2$ is greater than $1$ In factoring quadratics where the coefficient of $x^2$ is greater than $1$, I use the grouping method where we multiply the coefficient and constant together and then factor.
My question is can someone explain the math behind that?
Example:
$5x^2+11x+2,\quad 5\cdot2=10$
$5x^2+10x+x+2$
$5x(x+2)+1(x+2)$
$(5x+1)(x+2)$
|
Sometimes, instead of using symbols, it can help to just use numbers to keep track of patterns (and then generalize with symbols later).
To understand this grouping method, it is instructive to work the problem backwards, and to expand, instead of factor.
For example
$$(2x+3)(5x+7)\text{ the constants are all distinct primes. How fortunate.}$$
$$2\cdot5x^2 + 2\cdot7x+3\cdot5x+3\cdot7$$
$$\color{green}{2\cdot5}x^2 + (\color{blue}{2\cdot7}+\color{blue}{3\cdot5})x+\color{green}{3\cdot7}$$
Notice how we can partition the middle coefficient ($2\cdot7+3\cdot5$) into $2\cdot7$ and $3\cdot5$. The product of these partitions is $(2\cdot7)(3\cdot5)=2\cdot3\cdot5\cdot7$, which is equal to the product of the outside cofficients $(2\cdot5)(3\cdot7)=2\cdot3\cdot5\cdot7$.
In a real problem, however, we don't know the components (2, 3, 5, and 7). We only know the coefficients 10, 29, and 21 for the quadratic $$10x^2+29x+21$$
Our first step is to partition the middle coefficient into two parts such that the sum of the parts is $29$ and the product is $10\cdot21=210$. After analyzing the factors of $210$, we settle on the partitions $14$ and $15$.
Looking above at the quadratic with the colored coefficients, notice that neither of the factors of the $x^2$ term appear in both partitions (2 is a factor of the first partition, and 5 is a factor of the second partition). The same property holds true for the target factors of the constant term (the 3 and 7 appear in different partitions). This property will allow us to factor by grouping, thus discovering numbers 2,3,5, and 7 from the numbers 10, 14, 15, 21.
$$(10x^2+14x)+(15x+21)=2x(5x+7)+3(5x+7)=(2x+3)(5x+7)$$
Also take note that it doesn't matter what partition we associate with the $10x^2$ term. We just get a different common factor, depending upon whether we choose $14x$ or $15x$.
$$(10x^2+15x)+(14x+21)=5x(2x+3)+7(2x+3)=(5x+7)(2x+3)$$
As an exercise, you may want to restate the preceding argument for the quadratic
$$(px+r)(qx+s)$$
|
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|
Calculating point 2P on an elliptic curve The equation for the curve is $$y^2=x^3+ax+b$$ and the point in question is $P(x,y)$. We have to verify that the $x$ coordinate of $2P$ is $(x^4-2ax^2-8bx+a^2)/4y^2$. However, the value I get is $(9x^4+6ax^2-8xy^2+a^2)/4y^2$. I derived the algebraic formula for $2P$ ($x(2P)=m^2-2x)$ and used it to calculate the answer.How do I arrive at the actual answer?
|
We have:
$$\tag 1 y^2 = x^3 + ax + b$$
To add a point, we have:
*
*$\lambda = \dfrac{3x_1^2 + a}{2 y_1} = \dfrac{3x^2 + a}{2y}$
*$v = y_1 - \lambda x_1 = y-\dfrac{3x^2 + a}{2y}x$
*$x_3 = \lambda^2-x-x = \left(\dfrac{3x^2 + a}{2y}\right)^2-2x = \dfrac{9x^4+6ax^2-8xy^2+a^2}{4y^2}$
*$y_3 = -(\lambda x + v)$ (Calculate this out)
Next, we can substitute $(1)$ into the numerator of $x_3$, yielding:
$$\dfrac{9x^4+6ax^2-8xy^2+a^2}{4y^2} = \dfrac{9x^4+6ax^2-8x(x^3+ax+b)+a^2}{4y^2} = \dfrac{x^4-2ax^2-8bx+a^2}{4y^2}$$
Of course, you could have also done the same substitution in the denominator, but they chose not to.
|
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|
Equality involving Taylor coefficients Considering the following series expansion
$$
\frac{1}{{1 - 2x - x^2 }} = \sum\limits_{n = 0}^\infty {a_n } x^n
$$
prove that $
\forall n,\,\exists m
$ such that $
a_n ^2 + a_{n + 1} ^2 = a_m
$
I tried proving this using the coefficients computed by Wolfram but it didn't really help.
Any ideas?
|
First find the zeros of $1-2x-x^2$:
$$x=\frac{-2\pm\sqrt{4+4}}2=1\pm\sqrt2\;.$$
Let $\alpha=1+\sqrt2$ and $\beta=1-\sqrt2$; then
$$(1-\alpha x)(1-\beta x)=1-(\alpha+\beta)x+\alpha\beta x^2=1-2x-x^2\;,$$
so you can decompose $\frac1{1-2x-x^2}$ into partial fractions with denominators $1-\alpha x$ and $1-\beta x$ as
$$\frac1{1-2x-x^2}=\frac{A}{1-\alpha x}+\frac{B}{1-\beta x}\;.$$
I’ll leave it to you to calculate $A$ and $B$. Then observe that
$$\begin{align*}
\frac{A}{1-\alpha x}+\frac{B}{1-\beta x}&=A\sum_{n\ge 0}\alpha^nx^n+B\sum_{n\ge 0}\beta^nx^n\\\\
&=\sum_{n\ge 0}\left(A\alpha^n+B\beta^n\right)x^n\;,
\end{align*}$$
so $a_n=A\alpha^n+B\beta^n$, and
$$a_n^2+a_{n+1}^2=\left(A\alpha^n+B\beta^n\right)^2+\left(A\alpha^{n+1}+B\beta^{n+1}\right)^2\;.\tag{1}$$
Expand $(1)$ and collect terms, and with a little work you can find $m$ (in terms of $n$) such that it equals
$$A\alpha^m+B\beta^m=a_m\;.$$
|
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|
How many 2m-permutations, consisting only of cycles of even length? How many 2m-permutations, consisting only of cycles of even length?
I have found this formula:
$$Q_2(n) =((2n − 1)!!)^2$$
but how it can be proven?
|
The combinatorial species of permutations consisting only of even cycles is given by
$$\mathfrak{P}(\mathfrak{C}_{=2}(\mathcal{Z}) + \mathfrak{C}_{=4}(\mathcal{Z}) +
\mathfrak{C}_{=6}(\mathcal{Z})+ \mathfrak{C}_{=8}(\mathcal{Z}) + \cdots).$$
This translates to the generating function
$$G(z) = \exp\left(\frac{z^2}{2} + \frac{z^4}{4} + \frac{z^6}{6} + \frac{z^8}{8} + \cdots\right)$$
which is
$$\exp\left(\frac{1}{2}\left(\frac{z^2}{1} + \frac{z^4}{2} + \frac{z^6}{3} + \frac{z^8}{4} + \cdots\right)\right)$$
or
$$\exp\left(\frac{1}{2}\log\frac{1}{1-z^2}\right)
= \sqrt{\frac{1}{1-z^2}}.$$
By the Newton binomial we thus have the closed form for $n=2m$
$$(2m)! [z^{2m}] \sqrt{\frac{1}{1-z^2}}
= ((2m-1)!!)^2.$$
This is OEIS A001818.
Remark. The coefficient extraction can be performed by Lagrange inversion.
We have
$$[z^{2m}] G(z)
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{2m+1}} \sqrt{\frac{1}{1-z^2}} \; dz.$$
This is
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{(z^2)^{m+1}} \sqrt{\frac{1}{1-z^2}} \; z \; dz.$$
Put $1-z^2 = u^2 $ so that $- z \; dz = u\; du$ to get
$$- \frac{1}{2\pi i}
\int_{|u-1|=\epsilon} \frac{1}{(1-u^2)^{m+1}} \frac{1}{u} \times u \; du
\\ = - \frac{1}{2\pi i}
\int_{|u-1|=\epsilon} \frac{1}{(1-u)^{m+1}} \frac{1}{(1+u)^{m+1}} \; du
\\ = (-1)^m \frac{1}{2\pi i}
\int_{|u-1|=\epsilon} \frac{1}{(u-1)^{m+1}} \frac{1}{(2+u-1)^{m+1}} \; du
\\ = \frac{(-1)^m}{2^m} \frac{1}{2\pi i}
\int_{|u-1|=\epsilon} \frac{1}{(u-1)^{m+1}} \frac{1}{(1+(u-1)/2)^{m+1}} \; du
\\ = \frac{(-1)^m}{2^m} \frac{1}{2\pi i}
\int_{|u-1|=\epsilon} \frac{1}{(u-1)^{m+1}}
\sum_{q\ge 0} {q+m\choose m} (-1)^q \frac{(u-1)^q}{2^q} \; du
\\ = \frac{(-1)^m}{2^m} {2m\choose m} \frac{(-1)^m}{2^m}
= \frac{1}{2^{2m}} {2m\choose m}.$$
It follows that the answer is given by
$$(2m)! \times \frac{1}{2^{2m}} {2m\choose m}.$$
This implies that
$$\frac{Q_2(m)}{Q_2(m-1)}
= (2m)(2m-1) \frac{1}{2^2} \frac{(2m)(2m-1)}{m\times m} = (2m-1)^2$$
as pointed out by Robert Israel.
|
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|
Showing $\left(\frac{-1+\sqrt{3}i}{2}\right)^3=1$ In my textbook they asked me to show that
$$\left(\frac{-1+\sqrt{3}i}{2}\right)^3=1$$
but this is not true, I think.
I put down
$$\begin{align}
\frac12(-1+\sqrt{3}i)(-1+\sqrt{3}i)(-1+\sqrt{3}i)&=\frac12\left[(1-2\sqrt3i-3)(-1+\sqrt3i\right)]\\
&=\frac12\left[-1+2\sqrt3i+3+\sqrt3i+6-3\sqrt3i\right]\\
&=4\\
\end{align}$$
I checked it twice and I got $4$ what am I doing wrong?
|
$$\left(\frac{-1+\sqrt{3}i}{2}\right)^3=1$$
$$\frac{1}{2^3}(-1+\sqrt{3}i)(-1+\sqrt{3}i)(-1+\sqrt{3}i)=\frac12\left[(1-2\sqrt3i-3)(-1+\sqrt3i\right]$$
$$=\frac18\left[-1+2\sqrt3i+3+\sqrt3i+6-3\sqrt3i\right]$$
|
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|
Give a general formula for the following sequence I have a sequence where the terms are $1,0,-1,-1,0,1,1,0,-1,-1...$. Can someone help me find a general formula in terms of $n$ for this sequence?
|
We write the $n^{th}$ term of the sequence as $a_n$, then
$\displaystyle a_n = \begin{cases} 1 &\textrm{ if } n \equiv 1 \pmod 6 \\ 0 &\textrm{ if } n \equiv 2 \pmod 6 \\ -1 &\textrm{ if } n \equiv 3 \pmod 6 \\ -1 &\textrm{ if } n \equiv 4 \pmod 6 \\ 0 &\textrm{ if } n \equiv 5 \pmod 6 \\ 1 &\textrm{ if } n \equiv 0 \pmod 6 \end{cases}$
Consider the generating function of $a_n$, $\displaystyle f(x) = \sum\limits_{n=1}^{\infty} a_n x^n$
Then, $\displaystyle f(x) = \sum\limits_{n=0}^{\infty} (x^{6n+1}-x^{6n+3}-x^{6n+4}+x^{6n+6}) = \frac{x-x^3-x^4+x^6}{1-x^6} $
$\displaystyle = -1+ \frac{1}{1-x+x^2}$
Now, the roots of $1-x+x^2$ are $\alpha = \cos \frac{\pi}{3} + i\sin\frac{\pi}{3}$ and $\beta = \cos \frac{\pi}{3} - i\sin\frac{\pi}{3}$
Thus, $\displaystyle \frac{1}{1-x+x^2} = \frac{1}{\alpha - \beta}\left(\frac{1}{x-\alpha} - \frac{1}{x-\beta}\right) = \frac{2}{\sqrt{3}}\Im \left(\frac{\alpha}{1-\alpha x}\right) $
$\displaystyle = \sum\limits_{n=0}^{\infty} \frac{2}{\sqrt{3}}\sin \frac{(n+1)\pi}{3}x^n$
Thus, $\displaystyle a_n = \frac{2}{\sqrt{3}}\sin \frac{(n+1)\pi}{3}$
|
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|
Evaluate $\int_0 ^{\pi}\left (\frac{\pi}{2} - x\right)\sin\left(\frac{3x}{2}\right)\csc\left(\frac{x}{2}\right) dx$ How would you evaluate the integral
$$\int_0 ^{\pi} \left(\frac{\pi}{2} - x\right)\sin\left(\frac{3x}{2}\right)\csc\left(\frac{x}{2}\right) dx$$
The answer from Wolfram is $0$.
Would you use a substitution or do it by parts? Would making the substitution $u=\dfrac x2$ help?
|
Here are the steps
$$ \int_0 ^{\pi} \left(\frac{\pi}{2} - x\right)\sin\left(\frac{3x}{2}\right)\csc\left(\frac{x}{2}\right) dx $$
$$= \int_0 ^{\pi} \left(\frac{\pi}{2} - x\right)\left(\frac{\sin\left(\frac{3x}{2}\right)}{\sin\left(\frac{x}{2}\right)}\right) dx $$
$$= \int_0 ^{\pi} \left(\frac{\pi}{2} - x\right)\left(\frac{3\sin\left(\frac{x}{2}\right)-4\sin^3\left(\frac{x}{2}\right)}{\sin\left(\frac{x}{2}\right)}\right) dx $$
$$= \int_0 ^{\pi} \left(\frac{\pi}{2} - x\right)\left(3-4\sin^2\left(\frac{x}{2}\right)\right) dx $$
$$= \int_0 ^{\pi} \left(\frac{\pi}{2} - x\right)\left(3-4\left(\frac{1-\cos(x)}{2}\right)\right) dx $$
$$= \int_0 ^{\pi} \left(\frac{\pi}{2} - x\right)\left(3-2\left(1-\cos x\right)\right) dx $$
$$= \int_0 ^{\pi} \left(\frac{\pi}{2} - x\right)\left(1+2\cos x\right) dx $$
$$= \frac{\pi}{2}\int_0 ^{\pi}dx+\pi \int_0 ^{\pi}\cos x\ dx -\int_0^{\pi}x\ dx -2\int_0^{\pi}x\cos x\ dx $$
$$= \frac{\pi}{2}\bigg[x\bigg]_0 ^{\pi}+\pi \bigg[\sin x\bigg]_0 ^{\pi}-\bigg[\frac{x^2}{2}\bigg]_0^{\pi} -2\bigg[x\sin x+\cos x\bigg]_0^{\pi} $$
$$= \frac{\pi^2}{2}+\pi\sin \pi-\frac{\pi^2}{2}-2\bigg(\pi\sin \pi+\cos \pi-1\bigg) $$
$$= \pi\sin \pi-2\bigg(\pi\sin \pi+\cos \pi-1\bigg) $$
$$= 0-2\bigg(0-1-1\bigg) = -2\bigg(-2\bigg) =4$$
|
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|
Area of Triangle The position vectors of $A$ $B$ and $C$ relative to an origin $O$ are given by $OA=(2,1,3)$
$OB=(0,-1,7)$ and $OC=(2,4,7)$
Part i) Show that angle $BAC= \cos^{-1}(\frac{1}{3})$
Part ii) Using the previous part calculate the exact area of the triangle $ABC$
I was able to do both parts easily. For part two I used the formula $A=\frac{1}{2}ab\sin c$ and got the answer $\frac{20 \sqrt3}{3}=11.547...$
Then just out of curiosity I decided to use Heron's formula to see if I would get the same answer. From Heron's formula I got $5\sqrt5=11.1803...$
My Question: Why do I get different answers?
|
Due to the Cosine Theorem:
$$\cos A = \frac{b^2+c^2-a^2}{2bc}$$
and since, through the Pythagorean Theorem:
$$c^2 = \|OA-OB\|^2 = 2^2+2^2+4^2 = 24,$$
$$b^2 = \|OA-OC\|^2 = 0^2+3^2+4^2 = 25,$$
$$a^2 = \|OB-OC\|^2 = 2^2+5^2+0^2 = 29,$$
we have:
$$\cos A = \frac{25+24-29}{10\sqrt{24}}=\frac{1}{\sqrt{6}}$$
so:
$$\sin A = \sqrt{\frac{5}{6}}$$
and:
$$ [ABC]=\frac{1}{2}bc\sin A = \frac{1}{2}5\sqrt{24}\sqrt{\frac{5}{6}}=5\sqrt{5}.$$
|
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|
Indefinite integral with sector of ellipse An ellipse is given by the following equation:
$$
152 x^2 - 300 x y + 150 y^2 - 42 x + 40 y + 3 = 0
$$
After solving for the midpoint we have:
$$
152 (x-1/2)^2 - 300 (x-1/2) (y-11/30) + 150 (y-11/30)^2 = 1/6
$$
Introducing polar coordinates:
$$
x = 1/2 + r \cos(\theta) \quad ; \quad y = 11/30 + r \sin(\theta)
$$
Giving:
$$
152\, r^2 \cos^2(\theta) - 300\, r^2\, \cos(\theta) \sin(\theta) +
150\, r^2 \sin^2(\theta) = 1/6 \quad \Longrightarrow \\
\frac{1}{2} r^2(\theta) = \frac{1/12}
{152 \cos^2(\theta) - 300 \cos(\theta) \sin(\theta) + 150 \sin^2(\theta)}
$$
The area of a sector of the ellipse is:
$$
\int_{\theta_1}^{\theta_2} \frac{1}{2} r^2(\theta) \, d\theta
$$
So it seems that we have to find the indefinite integral:
$$
\int \frac{1/12 \, d\theta}
{152 \cos^2(\theta) - 300 \cos(\theta) \sin(\theta) + 150 \sin^2(\theta)}
$$
And then I'm stuck. Because feeding this into MAPLE with
int(1/12/(152*cos(theta)^2-300*cos(theta)*sin(theta)+150*sin(theta)^2),theta);
quite to my surprise, gives a complex result:
$$
{\frac {1}{720}}\,i\sqrt {3}\ln \left( \left( \tan \left( 1/2\,\theta
\right) \right) ^{2}+ \left( {\frac {5}{38}}\,i\sqrt {3}+{\frac {75}
{38}} \right) \tan \left( 1/2\,\theta \right) -1 \right) \\ -{\frac {1}{720
}}\,i\sqrt {3}\ln \left( \left( \tan \left( 1/2\,\theta \right)
\right) ^{2}+ \left( -{\frac {5}{38}}\,i\sqrt {3}+{\frac {75}{38}}
\right) \tan \left( 1/2\,\theta \right) -1 \right)
$$
But I'm pretty sure that the area of an ellipse sector is a real number.
So the question is: does there exist a closed form for the abovementioned
integral that is real valued instead of complex? What is it? And why that complex result with MAPLE?
AfterMath (see accepted answer)
$$
A = (1/2,3/10) \quad ; \quad B = (1/2,1/3) \quad ; \quad C = (1/2,11/30) \\
P = (1/3,1/6) \quad ; \quad Q = (2/3,1/2) \\
R = (1/4,1/10) \quad ; \quad S = (3/4,3/5)
$$
Triangle edges are black; ellipse is $\color{red}{red}$ .Triangle areas:
$\Delta PAQ = 1/180$ , $\Delta PRB = \Delta QSB = 1/720$ .
It is conjectured that the ellipse sector area $\overline{CRBSC}$
is exactly $1/3$ of the total ellipse area;the latter being
$ = \pi\sqrt{3}/180$ . Can someone prove or disprove this conjecture?
|
$$\int\frac{dx}{a\cos^2x-b\cos x\sin x+c\sin^2x}=\frac2{\sqrt\Delta}\cdot\tanh^{-1}\bigg(\frac{b-2c\tan x}{\sqrt\Delta}\bigg)$$ where $\Delta=b^2-4ac$. If $\Delta<0$, just use Euler's formula to transform the hyperbolic arctangent of complex argument into a trigonometric one of real argument. For $\Delta=0\iff b=\pm2\sqrt{ac}$,
we get $I=\dfrac{b+(a+c)\sin2x}{(a+c)\Big[(a-c)+(a+c)\cos2x\Big]}$.
|
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|
How do I prove this seemingly simple trigonometric identity $$a = \sin\theta+\sin\phi\\b=\tan\theta+\tan\phi\\c=\sec\theta+\sec\phi$$
Show that, $8bc=a[4b^2 + (b^2-c^2)^2]$
I tried to solve this for hours and have gotten no-where. Here's what I've got so far :
$$
\\a= 2\sin(\frac{\theta+\phi}{2})\cos(\frac{\theta-\phi}{2})
\\ b = \frac{2\sin(\theta+\phi)}{\cos(\theta+\phi)+\cos(\theta-\phi)}
\\c=\frac{2(\cos\theta+\cos\phi)}{\cos(\theta+\phi)+\cos(\theta-\phi)}
\\a^2 = \frac{\sin^2(\theta+\phi)[\cos(\theta+\phi)+1]}{\cos(\theta+\phi)+1}\\\cos(\theta-\phi)=\frac{ca}{b}-1\\\sin^2(\frac{\theta+\phi}{2})=\frac{2a^2b}{4(ca+b)}$$
|
I'd use the tangent half-angle substitution:
\begin{align*}
t&=\tan\frac\theta2 &
\sin\theta&=\frac{2t}{1+t^2} &
\tan\theta&=\frac{2t}{1-t^2} &
\sec\theta&=\frac{1+t^2}{1-t^2}
\\
u&=\tan\frac\phi2 &
\sin\phi&=\frac{2u}{1+u^2} &
\tan\phi&=\frac{2u}{1-u^2} &
\sec\phi&=\frac{1+u^2}{1-u^2}
\end{align*}
Then you have
\begin{align*}
a &= \frac{2 t^{2} u + 2 t u^{2} + 2 t + 2 u}{t^{2} u^{2} + t^{2} + u^{2} + 1} \\
b &= \frac{-2 t^{2} u - 2 t u^{2} + 2 t + 2 u}{t^{2} u^{2} - t^{2} - u^{2} + 1} \\
c &= \frac{-2 t^{2} u^{2} + 2}{t^{2} u^{2} - t^{2} - u^{2} + 1}
\end{align*}
\begin{multline*}
8bc=a[4b^2 + (b^2-c^2)^2] =\\
32\frac{t^{4} u^{3} + t^{3} u^{4} - t^{3} u^{2} - t^{2} u^{3} - t^{2} u - t u^{2} + t + u}{t^{4} u^{4} - 2 t^{4} u^{2} - 2 t^{2} u^{4} + t^{4} + 4 t^{2} u^{2} + u^{4} - 2 t^{2} - 2 u^{2} + 1}
\end{multline*}
The great benefit here is that after that first step, choosing to use that formulation, you almost don't have to think at all any more. From there on it's a straight-forward computation in rational functions. I left that to my computer algebra system, but if you are careful you can certainly do it by hand as well. The fact that my CAS expanded all those polynomials automatically makes the formulas above perhaps look more complicated than they really are, haven't tried.
|
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|
Closed form of $\int_{0}^{1}\frac{dx}{(x^2+a^2)\sqrt{x^2+b^2}}$ Is it possible to get a closed form of the following integral
$$\Phi(a,b)=\int_{0}^{1}\frac{dx}{(x^2+a^2)\sqrt{x^2+b^2}}$$
|
Here are a set of complete solutions, with all cases accounted for.
$$\Phi(a,b)=\int_{0}^{1}\frac{dx}{(x^2+a^2)\sqrt{x^2+b^2}}
\overset{x=\frac{bt}{\sqrt{1-t^2}} }=\int_0^{\frac1{\sqrt{1+b^2}} }\frac{dt}{a^2+(b^2-a^2)t^2}
$$
which leads to the solutions for the three distinctive cases below
\begin{align}
& \Phi(b<a) = \frac{1}{a\sqrt{a^2-b^2}}\tanh^{-1}\frac{\sqrt{a^2-b^2}}{a\sqrt{1+b^2}}\\
& \Phi(b=a) = \frac{1}{a^2\sqrt{1+a^2}}\\
& \Phi(b>a) = \frac{1}{a\sqrt{b^2-a^2}}\tan^{-1}\frac{\sqrt{b^2-a^2}}{a\sqrt{1+b^2}}\\
\end{align}
|
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|
Equations with variable powers Find the roots of the equation $3^{x+2}$+$3^{-x}$=10 . By inspection the roots are $x=0$ and $x=-2$. But how can I solve this equation otherwise?
|
Hint: Multiply both sides by the positive quantity $3^x$ to clear the fraction implied by the negative exponent, and rearrange a bit:
$$3^{x+2} + 3^{-x}=10$$
$$3^x\cdot3^{x+2} + 3^x\cdot 3^{-x} = 3^x\cdot 10$$
$$3^{2x+2} + 1 = 3^x\cdot 10$$
$$3^{2x}\cdot 3^2 + 1 = 3^x\cdot 10$$
$$(3^x)^2\cdot 9 + 1 = 3^x\cdot 10$$
$$9\cdot(3^x)^2 - 10\cdot(3^x) + 1 = 0$$
Now think of this as a quadratic equation in $3^x$:
$$9u^2 - 10u + 1 = 0$$
where $u = 3^x$.
Solve the quadratic to get the value(s) of $u$, say $u=A$ and $u=B$. Then remember that $u$ stands for $3^x$, so now you have to solve the equations $3^x=A$ and $3^x=B$, where $A$ and $B$ are the solutions to your quadratic.
Note that you will only have real solutions if $A>0$ or $B>0$, since $3^x$ can never be negative or zero if $x$ is real.
|
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|
Prove that $1 + 4 + 7 + · · · + 3n − 2 = \frac {n(3n − 1)}{2}$ Prove that
$$1 + 4 + 7 + · · · + 3n − 2 =
\frac{n(3n − 1)}
2$$
for all positive integers $n$.
Proof: $$1+4+7+\ldots +3(k+1)-2= \frac{(k + 1)[3(k+1)+1]}2$$
$$\frac{(k + 1)[3(k+1)+1]}2 + 3(k+1)-2$$
Along my proof I am stuck at the above section where it would be shown that:
$\dfrac{(k + 1)[3(k+1)+1]}2 + 3(k+1)-2$ is equivalent to $\dfrac{(k + 1)[3(k+1)+1]}2$
Any assistance would be appreciated.
|
The base case is trivial, now we follow to the inductive step by asuming the induction hypothesis and proving for $n + 1$.
So:
\begin{align*}
1 + 4 + 7 + ... + 3n-2 + 3(n+1)-2 & = \frac{n(3n-1)}{2} + 3(n+1)-2\\
& = \frac{n(3n-1)}{2} + \frac{2(3(n+1)-2)}{2}\\
& = \frac{3n^{2}-n+6n+6-4}{2}\\
& = \frac{3n^{2}+5n+2}{2}\\
& = \frac{(n+1)(3n+2)}{2}\\
& = \frac{(n+1)(3(n+1)-1)}{2}
\end{align*}
And we are done.
The important thing is to know when to apply the induction.
|
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|
Concise induction step for proving $\sum_{i=1}^n i^3 = \left( \sum_{i=1}^ni\right)^2$ I recently got a book on number theory and am working through some of the basic proofs. I was able to prove that $$\sum_{i=1}^n i^3 = \left( \sum_{i=1}^ni\right)^2$$ with the help of the identity $$\sum_{i=1}^ni = \frac{n(n+1)}{2}$$ My proof of the induction step goes as follows (supposing equality holds for all $k \in \{1,2,\dots n \})$: $$\sum_{i=1}^{n+1} i^3 = \sum_{i=1}^{n} i^3+(n+1)^3 \\ = \left( \sum_{i=1}^ni\right)^2+(n+1)^3 \\ = \left( \frac{n(n+1)}{2}\right)^2 + (n+1)^3 \\ = \frac{n^2(n+1)^2}{4}+\frac{4(n+1)^3}{4} \\ = \frac{n^4+2n^3+n^2}{4}+\frac{4n^3+12n^2+12n+4}{4} \\ = \frac{n^4+6n^3+13n^2+12n+4}{4} \\ = \frac{(n^2+3n+2)^2}{4} \\ = \frac{[(n+1)(n+2)]^2}{4} \\ = \left(\frac{(n+1)(n+2)}{2}\right)^2 \\ = \left(\sum_{i=1}^{n+1}i\right)^2$$ I was a little disappointed in my proof because the algebra got really hairy. It took me a long while to see that I could twice unfoil the polynomial $n^4+6n^3+13n^2+12n+4$ and all in all the solution seems pretty inelegant. Does anyone have a smoother way to prove the induction step or bypass the algebra? I feel like their must be some concise way to get the same result.
|
This is indeed possible. Starting from your fourth line: $$\frac{n^2(n+1)^2}{4} + \frac{4(n+1)^3}{4} = \frac{n^2(n+1)^2 + 4(n+1)(n+1)^2}{4} = \frac{(n^2 + 4(n+1))(n+1)^2}{4}$$ $$=\frac{(n^2 + 4n + 4)(n+1)^2}{4} = \frac{(n+2)^2(n+1)^2}{4}. $$
|
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|
Writing an alternating series as two non-alternating series? How does one calculate $$\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2}$$
if given this information: $$\sum_{n=1}^{\infty} \frac1{n^2} = \frac{\pi^2}{6}.$$
How does one account for the $(-1)^{n+1}$ in the first series?
Note: in this earlier post, user David Hodden (thanks) pointed out that the following simplification can be made:
$$\sum_{n=1}^{\infty} (-1)^{n+1}\frac1{n^2} = \sum_{n=1}^{\infty} \frac1{n^2} -2 \cdot \frac14 \sum_{n=1}^{\infty} \frac1{n^2} \\
= \frac{\pi^2}{12}$$
What I do not understand is where this simplification came from. Why can the alternating series be broken up into two non-alternating series?
|
There are two kind of terms in series $\displaystyle \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2}$: terms with sign $+$ and terms with sign $-$. Note that:
$$\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2}=\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{3^2}-\frac{1}{4^2}+\cdots$$
So $+$ occurs with terms for odd $n$ and $-$: for even terms. So let's divide this sum into two sums: for even and odd $n$:
$$\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2}=\sum_{n=1 \text{n is odd}}^{\infty} \frac{(-1)^{n+1}}{n^2}+\sum_{n=1 \text{n is even}}^{\infty} \frac{(-1)^{n+1}}{n^2}$$
You know that $-$ occurs for odd $n$ and $+$ for even, so you can write down this equality in different way:
$$\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2}=-\sum_{n=1 \text{n is odd}}^{\infty} \frac{1}{n^2}+\sum_{n=1 \text{n is even}}^{\infty} \frac{1}{n^2}$$
Next you can write down $\sum_{n=1 \text{n is even}}^{\infty} \frac{1}{n^2}$ in different way:
$$\sum_{n=1 \text{n is even}}^{\infty} \frac{1}{n^2}=\sum_{n=1}^{\infty} \frac{1}{(2n)^2}=\frac{1}{4}\sum_{n=1}^{\infty} \frac{1}{n^2}=\frac{\pi^2}{24}$$
On the other hand:
$$\sum_{n=1 \text{n is odd}}^{\infty} \frac{1}{n^2}=\sum_{n=1}^{\infty} \frac{1}{n^2}-\sum_{n=1 \text{n is even}}^{\infty} \frac{1}{n^2}=\frac{\pi^2}{6}-\frac{1}{4}\sum_{n=1}^{\infty} \frac{1}{n^2}=\frac{\pi^2}{6}-\frac{\pi^2}{24}$$
|
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|
Convergence test for the series $\sum_{n=1}^{\infty} \frac{\sqrt{n^2+1}-n}{\sqrt{n}}$
Determine convergence of the series
$$\sum_{n=1}^{\infty} \frac{\sqrt{n^2+1}-n}{\sqrt{n}}$$
My proof: using comparison test I have
$$\frac{\sqrt{n^2+1}-n}{\sqrt{n}} = \sqrt{n+\frac{1}{n}} - \sqrt{n} \to 0$$ for large $n$.
Therefore the series should be converges. Now that we need to show that $a_n \lt b_n$ for some $a_n$ s.t $\sum_{n=1}^{\infty} a_n$ converges. for all $n \in\mathbb N$,
$\frac{\sqrt{n^2+1}-n}{\sqrt{n}} \lt \frac{1}{n^2}$ since $\sum_{n=1}^{\infty} \frac{1}{n^2}$ converges then $\sum_{n=1}^{\infty} \frac{\sqrt{n^2+1}-n}{\sqrt{n}}$ must be converges.
Is this right or I need to fix my $\frac{1}{n^2}$?
|
Hint: $\dfrac{\sqrt{n^2+1} - n}{\sqrt{n}} = \dfrac{1}{\sqrt{n}\cdot \left(\sqrt{n^2+1} + n\right)} < \dfrac{1}{n^{3/2}}$
|
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|
The minimum value of $\frac{a(x+a)^2}{\sqrt{x^2-a^2}}$ The problem is to find the minimum of $A$, which I attempted and got a different answer than my book:
$$A=\frac{a(x+a)^2}{\sqrt{x^2-a^2}}$$ where $a$ is a constant
$A'=\frac{(x^2-a^2)^{\frac{1}{2}}(2a)(x+a)-\frac{1}{2}(x^2-a^2)^{-\frac{1}{2}}(a)(x+a)^2}{(x+a)(x-a)}$
$A'=\frac{(x^2-a^2)^{\frac{1}{2}}(2a)-\frac{1}{2}(x^2-a^2)^{-\frac{1}{2}}(a)(x+a)}{(x-a)}$
$0=(x^2-a^2)^{\frac{1}{2}}(2a)-\frac{1}{2}(x^2-a^2)^{-\frac{1}{2}}(a)(x+a)$
$(x^2-a^2)^{\frac{1}{2}}(2a)=\frac{1}{2}(x^2-a^2)^{-\frac{1}{2}}(a)(x+a)$
$2a\sqrt{x^2-a^2}=\frac{a(x+a)}{2\sqrt{x^2-a^2}}$
$2a(x+a)(x-a)=a(x+a)$
$2(x-a)=a$
$2x-2a=a$
$x=\frac{3}{2}a$
My book says x=2a. I'm not sure where I went wrong. Thanks in advance
|
Note
$$A'=\frac{(x^2-a^2)^{\frac{1}{2}}(2a)(x+a)-\frac{1}{2}\color{#C00000}{\cdot 2x}(x^2-a^2)^{-\frac{1}{2}}(a)(x+a)^2}{(x+a)(x-a)}$$
And Not:
$$A'=\frac{(x^2-a^2)^{\frac{1}{2}}(2a)(x+a)-\frac{1}{2}(x^2-a^2)^{-\frac{1}{2}}(a)(x+a)^2}{(x+a)(x-a)}$$
This is because:
$$\frac{d}{dx} {f(x)}^n = n (f(x))^{n-1}\color{green}{f'(x)}$$
So,
$$\frac{d}{dx} (x^2-a^2)^{\frac{1}{2}}=\frac{1}{2} (x^2-a^2)^{-\frac{1}{2}} \color{green}{2x} $$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluating $\int \frac{\sec^2 x}{(\sec x + \tan x )^{{9}/{2}}}\,\mathrm dx$ How do I evaluate
$$\int \frac{\sec^2 x}{(\sec x + \tan x )^{{9}/{2}}}\,\mathrm dx$$
I've started doing this problem by taking $u=\tan x$.
|
$\bf{My\; solution::}$ Given $$\displaystyle \int\frac{\sec^2 x}{(\sec x+\tan x)^{\frac{9}{2}}}dx$$
Let $$(\sec x+\tan x)= t\;,$$ Then $$\left(\sec x\cdot \tan x+\sec^2 x\right)dx = dt $$
So $$\displaystyle \left(\sec x+\tan x\right)dx = \frac{dt}{\sec x}\Rightarrow \displaystyle dx = \frac{dt}{\sec x}$$
Now Using $$\bullet \displaystyle (\sec^2 x-\tan^2 x) = 1\Rightarrow (\sec x+\tan x)\cdot (\sec x-\tan x) = 1$$
So $$\displaystyle (\sec x-\tan x) = \frac{1}{t}$$ and we above substitute $$\displaystyle (\sec x+\tan x)=t$$
So we get $$\displaystyle \sec x = \frac{1}{2}\left(t+\frac{1}{t}\right)$$
Now our Integral convert into $$\displaystyle \frac{1}{2}\int\frac{\sec x}{t^{\frac{9}{2}}}dt = \frac{1}{2}\int \frac{t+\frac{1}{t}}{t^{\frac{9}{2}}}dt = \frac{1}{2}\int t^{-\frac{7}{2}}dt+\frac{1}{2}\int t^{-\frac{11}{2}}dt$$
So we get $$\displaystyle \frac{1}{2}\cdot -\frac{2}{5}\cdot t^{-\frac{5}{2}}+\frac{1}{2}\cdot -\frac{2}{9}\cdot t^{-\frac{9}{2}}+\mathcal{C}$$
So $$\displaystyle \int\frac{\sec^2 x}{(\sec x+\tan x)^{\frac{9}{2}}}dx = -\frac{1}{5}\cdot (\sec x+\tan x)^{-\frac{5}{2}}-\frac{1}{9}\cdot (\sec x+\tan x)^{-\frac{9}{2}}+\mathcal{C}$$
|
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|
Why is the extension $k(x,\sqrt{1-x^2})/k$ purely transcendental? Consider the function field $k(x,\sqrt{1-x^2})$ of the circle over an algebraically closed field $k$. Is $k(x,\sqrt{1-x^2})$ a purely transcendental extension over $k$?
I'm curious because I was reading the answer here. The proof shows $k(x+\sqrt{1-x^2})=k(x,\sqrt{1-x^2})$. However, there is line which says,
$$
(x+\sqrt{1-x^2})^2 = x^2 + 2 \sqrt{1 - x^2} + (1 - x^2) = 2 \sqrt{1-x^2} + 1.
$$
But I think
$$
(x+\sqrt{1-x^2})^2 = x^2 + 2x\sqrt{1 - x^2} + (1 - x^2)=2x\sqrt{1-x^2}+1
$$
so, assuming $\operatorname{char}(k)\neq 2$, I believe at most one can conclude is $x\sqrt{1-x^2}\in k(x+\sqrt{1-x^2})$. I've been struggling to salvage this.
How can you show $k(x,\sqrt{1-x^2})$ is a purely transcendental extension over $k$? Is the extension still purely transcendental when $\operatorname{char}(k)=2$?
|
You are correct that there was a mistake in my original proof, but the result is still true. Instead, in the case that $\mathrm{char}(k) \neq 2$, consider the purely transcendental extension $k \left( \frac{\sqrt{1-x^2}}{1+x} \right)$. This field contains
$$
\frac{1 - \left(\frac{\sqrt{1-x^2}}{1+x}\right)^2}{1 + \left(\frac{\sqrt{1-x^2}}{1+x}\right)^2} = \frac{(1+x)^2 - (1-x^2)}{(1+x)^2 + (1-x^2)} = x
$$
From this it is easy to see that $x, \sqrt{1-x^2} \in k \left( \frac{\sqrt{1-x^2}}{1+x} \right)$, hence $k(x, \sqrt{1-x^2}) = k \left( \frac{\sqrt{1-x^2}}{1+x} \right)$.
Really this all just comes from the rational parametrization of the circle $t \mapsto \left( \frac{1 - t^2}{1+t^2}, \frac{2t}{1+t^2} \right)$.
In the case that $\mathrm{char}(k) = 2$, the extension $k(x)$ works, as
$$(1+x)^2 = 1 + x^2 = 1- x^2,$$
hence $\sqrt{1-x^2} \in k(x)$. This case is much easier because the circle $x^2 + y^2 + 1 = 0$ is really just a degenerate double line $(x+y+1)^2 = 0$ in characteristic 2.
|
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|
Find the number of all 3 digit numbers $n$ such that $S(S(n))=2$ For any natural number $n$ ,let $S(n)$ denote the sum of the digits of $n$.Find the number of all 3 digit numbers $n$ such that $S(S(n))=2$
|
Write $n=abc=100a+10b+c.$ Then, $S(n)=a+b+c.$ Note that $S(n)$ has two digits. Then,
if $a+b+c<10$ then $a+b+c=2$ (since $S(S(n))=a+b+c);$
if $10\le a+b+c<20$ then $a+b+c=11$ (since $S(S(n))=a+b+c-9);$
if $20\le a+b+c$ then $a+b+c=20$ (since $S(S(n))=a+b+c-18).$
|
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|
The minimum range of a $n \times n$ grid Let $n > 1$ be a integer, we put $1, 2, \cdots n^2$ into the cells of a $n \times n$ grid.
Let the range of the grid be the maximal difference between two cells that are in the same row or in the same column. Find the minimum range of all possible grids.
|
For $n$ even let $n=2k$.
$\left(\begin{array}{cccccc}
1&\cdots & k^2-k + 1&k^2+1&\cdots&2k^2-k+1\\
\vdots & \ & \vdots& \vdots&\ &\vdots\\
k &\cdots & k^2 & k^2+k& \cdots \ & 2k^2\\
n^2-2k^2+1&\cdots&n^2-k^2-k+1&n^2-k^2+1&\cdots&n^2-k+1\\
\vdots&\ &\vdots&\vdots&\ &\vdots\\
n^2-2k^2+k&\cdots&n^2-k^2&n^2-k^2+k&n^2-k&n^2\\
\end{array} \right)$
is an optimal solution. ($\cdots$) stands for "increasing by $k$" and ($\ \vdots\ $) for "increasing by $1$".
All rows and columns are increasing. We therefore find the range by considering the differences of the last and first entry in each row and column. The maximum for this is $n^2-2k^2+k-1$, or in terms of $n$: $$\frac{n^2}{2}+\frac n 2 -1.$$
Proof that this is optimal: We are trying to place the smallest and the largest numbers in a way in which they interfere (i.e. are in the same row or column) with the least difference. This is done by putting the smallest $k^2$ numbers in the top left corner and the largest $k^2$ in the bottom right. The next number we have to place is $k^2+1$ (or equivalently $n^2-k^2$). You will see that this yields that the minimum difference in the row or column in which we place it is $n^2-2k^2+k-1$. By filling up the other fields, we obtain an optimal matrix.
For odd $n$ the lowest $\left(\frac{n+1}2\right)^2$ numbers can be put in the top left corner and the highest $\left(\frac{n-1}2\right)^2$ in the bottom right. This would give a minimum range of $$n^2-\left(\frac{n-1}{2}\right)^2-\left(\frac{n+1}{2}\right)^2+\frac{n-1}{2}=\frac 1 2 (n^2+n-2),$$
as this is the minimal difference between the lest element which does not fit into the bottom right corner any more and the elements in the top left corner it interferes with.
|
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|
Prove by induction that an expression is divisible by 11
Prove, by induction that $2^{3n-1}+5\cdot3^n$ is divisible by $11$ for any even number $n\in\Bbb N$.
I am rather confused by this question. This is my attempt so far:
For $n = 2$
$2^5 + 5\cdot 9 = 77$
$77/11 = 7$
We assume that there is a value $n = k$ such that $2^{3k-1} + 5\cdot 3^k$ is divisible by $11$.
We show that it is also divisible by $11$ when $n = k + 2$
$2^{3k+5} + 5\cdot 3^{k+2}$
$32\cdot 2^3k + 5\cdot 9 \cdot3^k$
$32\cdot 2^3k + 45\cdot 3^k$
$64\cdot 2^{3k-1} + 45\cdot 3^k$ (Making both polynomials the same as when $n = k$)
$(2^{3k-1} + 5\cdot 3^k) + (63\cdot 2^{3k-1} + 40\cdot 3^k)$
The first group of terms $(2^{3k-1} + 5\cdot 3^k)$ is divisible by $11$ because we have made an assumption that the term is divisible by $11$ when $n=k$. However, the second group is not divisible by $11$. Where did I go wrong?
|
Hint. Let $A_k = 2^{3k-1}+5\cdot 3^k = \frac{1}{2}\cdot 8^k+5\cdot 3^k$.
Once you prove that the sequence $\{A_k\}_{k\geq 1}$ fulfills the recurrence relation
$$ A_{k+2} = 11\cdot A_{k+1} -24\cdot A_k $$
you trivially have that $A_{k+2}\equiv -2\cdot A_k\pmod{11}$ holds.
$A_2=77\equiv 0\pmod{11}$ and the conclusion is straightforward.
|
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|
Limit of function using Taylor's Formula To find:
$$\lim_{x \to 0} \left(\frac{ \sin(x)}{x}\right)^\frac{1}{x}$$
by using Taylor's formula.
So I used the Taylor's formula for $\sin(x)$ and got::
$\sin(x) = x - \frac{x^3}{6} + O(x^4)$
And then my function becomes:
$$\lim_{x \to 0} \left(\frac{ x - \frac{x^3}{6} + O(x^2)}{x}\right)^\frac{1}{x} = \lim_{x \to 0} \left(1 + \frac{x^2}{6} + O(x)\right)^\frac{1}{x}$$
After that I lost. How am i suppose to get rid of the power of $\frac{1}{x}$??
|
$$
\lim_{x \to 0} \left( 1 + \frac {x^2}6\right )^{\frac 1x} = \lim_{x \to 0} \left[\left ( 1 + \frac {x^2}6\right )^{\frac 6{x^2}} \right ]^{\frac x6} = \lim_{x \to 0}e^{\frac x6} = 1
$$
|
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|
Evaluate this Trigonometric Expression: $\sqrt[3]{\cos \frac{2\pi}{7}} + \sqrt[3]{\cos \frac{4\pi}{7}} + \sqrt[3]{\cos \frac{6\pi}{7}}$
Evaluate
$$ \sqrt[3]{\cos \frac{2\pi}{7}} + \sqrt[3]{\cos \frac{4\pi}{7}} + \sqrt[3]{\cos \frac{6\pi}{7}}$$
I found the following
*
*$\large{\cos \frac{2\pi}{7}+\cos \frac{4\pi}{7} + \cos \frac{6\pi}{7}=-\dfrac{1}{2}}$
*$\large{\cos \frac{2\pi}{7}\times\cos \frac{4\pi}{7} + \cos \frac{4\pi}{7}\times\cos \frac{6\pi}{7} + \cos \frac{6\pi}{7}\times\cos \frac{2\pi}{7}}=-\dfrac{1}{2}$
*$\large{\cos \frac{2\pi}{7}\times\cos \frac{4\pi}{7}\times\cos \frac{6\pi}{7}=\dfrac{1}{8}}$
Now, by Vieta's Formula's, $\large{\cos \frac{2\pi}{7}, \cos \frac{4\pi}{7} \: \text{&} \: \cos \frac{6\pi}{7}}$ are the roots of the cubic equation
$$8x^3+4x^2-4x-1=0$$
And, the problem reduces to finding the sum of cube roots of the solutions of this cubic.
For that, I thought about transforming this equation to another one whose zeroes are the cube roots of the zeroes of this cubic by making the substitution
$$x\mapsto x^3$$
and getting another equation
$$8x^9+4x^6-4x^3-1=0$$
However, this new equation will have some extra roots too and we can't directly use Vieta's to get the desired sum.
Also, it's given that the sum evaluates to a radical of the form
$$\sqrt[3]{\frac{1}{d}(a-b\sqrt[b]{c})}$$
where $a, b, c \: \text{&} \: d \in \mathbb Z$
Can somebody please help me with this question?
Thanks!
|
Try using this:
$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$
Let $a^3= \cos{\frac{2\pi}{7}},b^3= \cos{\frac{4\pi}{7}},c^3= \cos{\frac{6\pi}{7}} $.
|
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|
Congruence properties of $a^5+b^5+c^5+d^5+e^5=0$? It is known that given a solution to,
$$a^4+b^4+c^4 = d^4\tag1$$
then either $-c+d,\;c+d$ is always divisible by $2^{10}$. For example,
$$95800^4+414560^4+217519^4=422481^4$$
then $217519+422481=2^{10}\cdot5^4$.
Duncan Moore noticed, but could not prove, a similar congruence for,
$$a^5+b^5+c^5+d^5+e^5=0\tag2$$
There are only three known primitive solutions $a,b,c,d,e$, namely,
$$27,\; 84,\; 110,\; 133,\; -144$$
$$220,\; -5027,\; -6237,\; -14068,\; 14132$$
$$55,\; 3183,\; 28969,\; 85282,\; -85359$$
And we have,
$$27 + 133 = \color{blue}{2^5}\cdot5$$
$$-5027 -6237 = -\color{blue}{2^{10}}\cdot 11,\quad \text{and}\quad -14068 + 14132 = \color{blue}{2^6}$$
$$55 + 28969 = \color{blue}{2^5}\cdot907,\quad \text{and}\quad 3183 + (- 85359) = -\color{blue}{2^8}\cdot327$$
Question: Is it true that solutions to $(2)$ always have a pair of addends such that $a+b$ is divisible by $2^5$?
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It is helpful to consider separately the cases with exactly two and exactly four odd terms (these are the only possibilities since a solution with zero odd terms would not be primitive, while a sum with an odd number of odd terms could not equal zero).
The case with exactly four odd terms looks difficult, but that with exactly two odd terms (which includes two of the three known solutions) is straightforward. Without loss of generality suppose $a, b$ in $(2)$ are odd and $c, d, e$ even. Then:
$$2^5\mid -(c^5+d^5+e^5)=(a^5 + b^5) = (a+b)(a^4-a^3b+a^2b^2-ab^3+b^4)$$
The final set of brackets contains five odd terms and is therefore odd. Hence:
$$2^5 \mid a+b$$
This result generalises to any number of terms, provided that exactly two terms are odd, and also to any odd power $n$, provided that we consider divisibility by $2^n$. An example for $n=7$ is:
$$194^7 + 150^7+105^7 +23^7 -192^7-152^7-132^7-38^7=0$$
where $105+23=2^7$.
|
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Differentiating $ \left( 1 - \frac {1}{x} \right)^x $ I have a calculus question. How does one differentiate $\left(1-\frac{1}{x}\right)^x$, for x>1? It should be positive right?
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Notice that the power is $x$, which is not a constant, so if we let $ y = \left( 1 - \frac {1}{x} \right)^x $, Then we can take the natural logarithm to get $$ \ln y = \ln \left( 1 - \frac {1}{x} \right)^x = x \cdot \ln \left( 1 - \frac {1}{x} \right). $$
Differentiating each side (implicit differentiation), we get $$ \begin {align*} \frac {y'}{y} &= x \cdot \frac {\left( 1 - \frac {1}{x} \right)'}{1 - \frac {1}{x}} + 1 \cdot \ln \left( 1 - \frac {1}{x} \right) \\&= x \cdot \frac {- \frac {1}{x^2}}{1 - \frac {1}{x}} + \ln \left( 1 - \frac {1}{x} \right) \\&= \frac {1}{1 - x} + \ln \left( 1 - \frac {1}{x} \right), \end {align*} $$so $ y' = \left( 1 - \frac {1}{x} \right)^x \cdot \left( \frac {1}{1-x} + \ln \left( 1 - \frac {1}{x} \right) \right) $.
Now, I don't know exactly what you're asking since it's unclear what the question is, but hopefully whatever it is, you can finish from here. In general, if $ y = \left[ u(x) \right]^x $, you can follow the same steps and find that $$ y' = \left[ u(x) \right]^{x-1} \cdot x \cdot u'(x) + \left[ u(x) \right]^x \cdot \ln \left( u(x) \right). $$
|
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Find all integers that make this expression rational I came up with this difficult problem a while ago while solving another relatively easy problem.
Find all integers m and n, such that $m^2 + n^2$ is a square, and such that $\sqrt{\frac{2m^2+2}{n^2+1}}$ is rational.
I've already tried the pythagorean substitution numerous times to no avail and to make things worse it isn't even homogenous. Come to think of it I must have made a mistake in reducing my problem to this.
|
Too long for a comment. As requested by OP as a sidenote, if we relax the requirements on $m,n$ and allow rationals, then the system,
$$m^2+n^2 = x^2,\quad \frac{2(m^2+1)}{n^2+1}=y^2\tag1$$
does have an infinite number of rational solutions. Other than the obvious case $n=1$, we have two quadratic polynomials in $m$ to be made squares,
$$f(m) = x^2,\quad\quad g(m) = y^2$$
hence an intersection of two quadric surfaces. A simple rational solution to $(1)$ is then,
$$m = \frac{(a-c)c}{(a-2c)b},\quad n=\frac{a}{2b}$$
where,
$$a^2+2b^2 = c^2\tag2$$
and $(2)$ is easily solved as $a,b,c = u^2-2v^2,\, 2uv,\, u^2+2v^2$ yielding,
$$x^2=\frac{(u^4+4u^2v^2+20v^4)^2}{(4uv)^2(u^2+6v^2)^2},\quad y^2 = \frac{2(u^2+4v^2)\,(4v)^2}{(u^2+6v^2)^2}$$
Thus, it remains to find $u^2+4v^2 = 2z^2$ which can also be solved parametrically. For a particular example, let $u,v,z = 2,\,1,\,4,$ so $m,n = \frac{3}{5},\frac{1}{4},$ and $x,y = \frac{13}{20},\frac{8}{5}$.
P.S. There are no positive integer solutions to $(1)$ with $m,n<500$.
|
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Greatest value of $f(x)= (x+1)^{1/3}-(x-1)^{1/3}$ on $(0,1)$ Greatest value of $f(x)= (x+1)^{1/3}-(x-1)^{1/3}$ on $(0,1)$
Please guide me to solve this problem. I have differentiated it with respect to $x$ and make equal to zero, but couldn't get any point.
|
HINT: Here we can use the trigonometric substitution $x=\cos 2\theta$ where $\theta\in[0,\frac{\pi}{4}]$
Now $$f(x)=(x+1)^{1/3}-(x-1)^{1/3}=(2\cos^2 2\theta)^{1/3}-(2\sin^2 2\theta)^{1/3}.$$ Then we can use the parametric differentiation for find the minimum and maximums.
OR:
Using the identity $a^3-b^3=(a-b)(a^2+ab+b^2)$ we can obtain that, $$(x+1)^{1/3}-(x-1)^{1/3}=\dfrac{2}{(x+1)^{2/3}+(x+1)^{1/3}(x-1)^{1/3}+(x-1)^{1/3}}.$$ Therefore it is enough, find the minimum value of $$a^2+ab+b^2=\Big(a+\dfrac{b}{2}\Big)^2+\dfrac{3b^2}{4}$$ in which $a=-\dfrac{b}2,$ where $a=(x+1)^{1/3},\,\,\,b=(x-1)^{1/3}$ and $x\in[0,1]$.
|
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Finding the residue of a function with a surd variable $$f(z) = \frac{1}{z-2\sqrt{z}+2}$$
Is this the correct way of doing this, please advice - Thanks, what i did was try to rationalise the expression first as follows:
$$f(z) = \frac{1}{z-2\sqrt{z}+2} = \frac{1}{(z+2)-2\sqrt{z}}*\frac{(z+2)+2\sqrt{z}}{(z+2)+2\sqrt{z}}$$
$$\therefore$$
$$f(z) = \frac{(z+2)+2\sqrt{z}}{z^2+4}$$
My poles are at $z= ±4i$
$$\operatorname{Res}(f(z)e^{zt}, 4i) = \lim_{z\to 4i} (z-4i)\frac{e^{zt}((z+2)+2\sqrt{z})}{(z-4i)(z+4i)}=\lim_{z\to 4i} \frac{e^{zt}(z+2+2\sqrt{z})}{(z+4i)}$$
$$\operatorname{Res}(f(z)e^{zt}, 4i) =\frac{e^{4it}(4i+2+2\sqrt{4i})}{8i}$$
and
$${Res}(f(z)e^{zt}, -4i) = \lim_{z\to -4i} (z+4i)\frac{e^{zt}((z+2)+2\sqrt{z})}{(z-4i)(z+4i)}=\lim_{z\to-4i} \frac{e^{zt}(z+2+2\sqrt{z})}{(z-4i)}$$
$$\operatorname{Res}(f(z)e^{zt}, -4i) =\frac{e^{-4it}(-4i+2+2i\sqrt{4i})}{-8i}$$
$$\sum \operatorname{Res}(f(z)e^{zt}, 4i)=\frac{e^{4it}(4i+2+2\sqrt{4i})}{8i}+\frac{e^{-4it}(-4i+2+2i\sqrt{4i})}{-8i}$$
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Why are calculated The function {f\left( x \right)e^x }
\begin{array}{l}
{\mathop{\rm Re}\nolimits} s\left( {f\left( x \right); - 2i} \right) = \mathop {\lim }\limits_{z \to - 2i} \left( {z + 2i} \right)\left( {f(z)} \right) \\
= \mathop {\lim }\limits_{z \to - 2i} \frac{{z + 2 + 2\sqrt z }}{{z - 2i}} = \frac{{ - 2i + 2 + 2\sqrt { - 2i} }}{{ - 4i}} \\
\end{array}
\begin{array}{l}
{\mathop{\rm Re}\nolimits} s\left( {f\left( x \right);2i} \right) = \mathop {\lim }\limits_{z \to - 2i} \left( {z - 2i} \right)\left( {f(z)} \right) \\
= \mathop {\lim }\limits_{z \to + 2i} \frac{{z + 2 + 2\sqrt z }}{{z + 2i}} = \frac{{ + 2i + 2 + 2\sqrt { + 2i} }}{{ + 4i}} \\
\end{array}
|
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How to evaluate $ \lim_{x \to 1} \frac{e^{x-1}-1}{x^2-1} $ How can you calculate this limit?
$$
\lim_{x \to 1} \frac{e^{x-1}-1}{x^2-1}
$$
I really don't have a clue what to do with the $e^{x-1}$
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The form of the expression in the limit is something like a difference quotient, and in fact, applying the difference-of-squares factorization $a^2 - b^2 = (a + b)(a - b)$ to the bottom shows that we can write the expression as the product of a difference quotient (whose limit is, by definition, a derivative) and a function continuous at the limit point $x = 1$:
$$\frac{e^{x - 1}}{x^2 - 1} = \color{#00bf00}{\frac{1}{x + 1}} \color{#bf0000}{\frac{e^{x - 1} - 1}{x - 1}}.$$
We can evaluate the limit of this expression by showing that the limits of each of the factors exist and computing them.
The $\color{#bf0000}{\text{red}}$ expression is the difference quotient of the function $$f(x) := e^{x - 1}$$ at $1$, so its limit is
$$\lim_{x \to 1} \color{#bf0000}{\frac{e^{x - 1} - 1}{x - 1}} = f'(1) = e^{x - 1}\vert_{x = 1} = 1.$$
The limit of the $\color{#00bf00}{\text{green}}$ expression is $$\lim_{x \to 1} \color{#00bf00}{\frac{1}{x + 1}} = \frac{1}{2}$$ and so
$$\lim_{x \to 1} \frac{e^{x - 1}}{x^2 - 1} =\left(\lim_{x \to 1} \color{#00bf00}{\frac{1}{x + 1}}\right) \left(\lim_{x \to 1} \color{#bf0000}{\frac{e^{x - 1} - 1}{x - 1}}\right) = \left(\frac{1}{2}\right)(1) = \frac{1}{2}.$$
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|
Trigonometic Substitution VS Hyperbolic substitution The following tables were taken from University of Pennsylvania's page about Calculus:
Trigonometric Substitution
Hyperbolic Substitution
As you can see, the forms $1+x^2$ and $x^2-1$ are repeated in the tables. How does one know when a trigonometric substitution is more suitable to a problem than a hyperbolic substitution and vice versa?
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In my opinion, the following substitutions are enough:
appears in integrand
substitution
convention
1
$\sqrt{x^2+1}$
$x=\sinh\theta$
$\theta\in\mathbb R$
2
$\sqrt{x^2-1}$
$x=\pm\cosh\theta$
$\theta\geq0$
3
$\sqrt{1-x^2}$
$x=\sin\theta$
$-\frac{\pi}2\leq\theta\leq\frac{\pi}2$
$\sinh$ and $\cosh$ are better substitutions than $\tan$ and $\sec,$ respectively, as they are easier to differentiate and integrate, and have nicer principal domains. $\sin$ is a better substitution than $\tanh$ as it is easier to differentiate and integrate. The following examples illustrate this:
integrand
substitution
result
1
$\displaystyle\frac{x^2}{\sqrt{x^2+4}}$
$x=2\sinh\theta\quad\color{green}✔$
$4\int\sinh^2\theta\,\mathrm d\theta$
$\displaystyle\frac{x^2}{\sqrt{x^2+4}}$
$x=2\tan\theta\quad\color{red}✗$
$4\int\tan^2\theta\sec\theta\,\mathrm d\theta$
2
$\displaystyle\frac{\sqrt{x^2-4}}{x^2}\quad(x\leq-2)$
$x=-2\cosh\theta\quad\color{green}✔$
$-\int\tanh^2\theta\,\mathrm d\theta$
$\displaystyle\frac{\sqrt{x^2-4}}{x^2}\quad(x\leq-2)$
$x=2\sec\theta\quad\color{red}✗$
$-\int\tan^2\theta\cos\theta\,\mathrm d\theta$
3
$\sqrt{9-x^2}$
$x=3\sin\theta\quad\color{green}✔$
$9\int\cos^2\theta\,\mathrm d\theta$
$\sqrt{9-x^2}$
$x=3\tanh\theta\quad\color{red}✗$
$9\int\mathrm{sech^3}\,\theta\,\mathrm d\theta$
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|
Series expansion of $\frac{1}{(1+x)(1−x)(1+x^2)(1−x^2)(1+x^3)(1−x^3)\cdots}$? How would I find the series expansion $\displaystyle\frac{1}{(1+x)(1−x)(1+x^2)(1−x^2)(1+x^3)(1−x^3)\cdots}$ so that it will turn into an infinite power series again??
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I suppose that you can make the expresion shorter using $(1+a)(1-a)=1-a^2$ so $$A=\frac{1}{(1+x)(1−x)(1+x^2)(1−x^2)(1+x^3)(1−x^3)\cdots}=\frac{1}{(1−x^2)(1−x^4)(1-x^6)\cdots}$$ and now use the fact that $$\frac{1}{1-y}=\sum_{i=0}^{\infty}y^i$$ and replace successively $y$ by $x^2$, $x^4\cdots$,$x^{2n}$ before computing the overall product.
Doing so, for a large number of terms, you should arrive to $$A=1+x^2+2 x^4+3 x^6+5 x^8+7 x^{10}+11 x^{12}+15 x^{14}+22 x^{16}+30 x^{18}+42 x^{20}+56
x^{22}+77 x^{24}+101 x^{26}+135 x^{28}+176 x^{30}+O\left(x^{31}\right)$$ and, as mentioned earlier, the coefficients correspond to sequence $\text{A00041}$ of $\text{EOIS}$ where $a_n$ is the number of partitions of $n$ (the partition numbers).
Probably off-topic, for an infinite number of terms, $$A=\frac{1}{\left(x^2;x^2\right){}_{\infty }}$$ where appears the q-Pochhammer symbol and which, for sure, leads to the same expansion.
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An integer when divided by $5$ and $13$ leaves residues $4$ and $7$ respectively Find an integer when divided by $5$ and $13$ leaves residues $4$ and $7$ respectively. (Without Modular Arithmetic).
I don't know if it is right, but I got this
$$n=5x+4=13y+7$$
$$n=5(x-11)+59=13(y-4)+59$$
The $\operatorname{lcd}(5,13)=65$
So our number is every number like $n=65k+59$
I'll be glad if you enlighten me it.
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$5x - 13y = 7-4 = 3 \Rightarrow 5x = 15y + 3 - 2y \Rightarrow x = 3y + \dfrac{3-2y}{5} \Rightarrow 5\mid 3-2y \Rightarrow 3-2y = 5k \Rightarrow y = \dfrac{3-5k}{2} = \dfrac{3-k}{2} - 2k \Rightarrow 2\mid 3-k \Rightarrow 3-k = 2h \Rightarrow k = 3-2h \Rightarrow 3 - 5k = 3 - 5(3-2h) = -12 + 10h \Rightarrow y = \dfrac{3-5k}{2} = \dfrac{-12+10h}{2} = 5h-6 \Rightarrow n = 13y+7 = 13(5h-6)+7 = 65h-71 = 65(h-1) - 6 = 65t-6 = 65(t-1) + 59=65m+59, m \in \mathbb{Z}$
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Integration without complex analysis on rational-improper integral Evaluate:
$$\int_{0}^{\infty} \frac{1}{x^6 + 1} \,\mathrm dx$$
Without the use of complex-analysis.
With complex analysis it is a very simple problem, how can this be done WITHOUT complex analysis?
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Let $\displaystyle \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\mathcal I=\int_0^\infty\frac{1}{1+x^6} \,\mathrm dx
$
$$\begin{align}
I&=\frac{1}{2}\left[ \int_0^\infty \frac{(1-x^2+x^4)+x^2+(1-x^4)}{(1+x^2)(1-x^2+x^4)} \,\mathrm dx \right]\tag{1}\\
&=
\frac{1}{2}\left[\int_0^\infty \frac{1}{1+x^2} \,\mathrm dx
+ \int_0^\infty \frac{x^2}{1+x^6} \,\mathrm dx
+ \color{grey}{\int_0^\infty \frac{1-x^2}{1-x^2+x^4} \,\mathrm dx}\right] \tag{2}\\
&=\frac{1}{2}\left[\frac\pi2+ \frac\pi{6} +\color\grey{0} \right] \tag{3}\\
&\mathcal I=\frac{\pi}{3} \tag{4}\\
\end{align}$$
$$\int_0^\infty\frac{1}{1+x^6} \,\mathrm dx=\frac\pi3$$
$\text{ Explanation : }(3)$
$$
\small\color\grey{J=\int_0^\infty \frac{1-x^2}{1-x^2+x^4} \,\mathrm dx}
=\int_0^1 \frac{1-x^2}{1-x^2+x^4} \,\mathrm dx
+\int_1^\infty \frac{1-x^2}{1-x^2+x^4} \,\mathrm dx$$
Now substitute
$\small\displaystyle x=\frac1t$ in second integral, To get
$$
\small\color\grey{J=\int_0^\infty \frac{1-x^2}{1-x^2+x^4} \,\mathrm dx}
=\int_0^1 \frac{1-x^2}{1-x^2+x^4} \,\mathrm dx
-\int_0^1 \frac{1-t^2}{1-t^2+t^4} \,\mathrm dt=\color\grey0$$
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Integrating $\int \frac{dx}{x^2+x+1}$ I am trying to evaluate the following integral:
$$I=\int \frac{dx}{x^2+x+1}$$
I am not supposed to do it with complex numbers so it's kind of hard.
I checked the answer on WolframAlpha.
It gives $$I=\frac{\sqrt{3}}{2}\arctan{\left(\frac{\sqrt{3}}{2}x+\frac{1}{2}\right)}+C $$
Having this information I found that I need to set
$$t=\frac{\sqrt{3}}{2}x+\frac{1}{2}$$
consequently: $x^2+x+1=\frac{3}{4}(t^2+1)$ and $dx=\frac{\sqrt{3}}{2}dt$
therefore: $$I=\frac{2}{\sqrt{3}} \int \frac{dt}{t^2+1}$$ which can be found easily by setting $t=\tan{\theta}$ giving the same answer stated above.
Practically, I cheated: if I had not known the final answer I would have not guessed what I needed to set as $t$.
My question is: how should I think in order to find this integral?
Is there a method to find integrals of the form $$\int \frac{dx}{ax^2+bx+c}$$
Without knowing the answer, how can I solve such integral?
Thanks
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Answering your second question.
Since $ax^2 + bx+c = a\Bigg[x^2 + \frac{b}{a}x + \frac{c}{a}\Bigg] = a\Bigg[(x+\frac{b}{2a})^2 - \frac{b^2 - 4ac}{4a^2} \Bigg]$ then
$$\int \frac{1}{ax^2+bx+c}dx = \frac{1}{a}\int \frac{1}{(x+\frac{b}{2a})^2 - \frac{b^2 - 4ac}{4a^2}}dx$$
And the next step will depend on the sign of $b^2-4ac$.
|
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|
Cube roots escape $ \sqrt{\sqrt[3]{5}-\sqrt[3]{4}} \times 3 = \sqrt[3]{a} + \sqrt[3]{b} - \sqrt[3]{c}, $
where $ a, b $ and $ c $ are positive integers. What is the value of $ a+b+c $?
This question appeared in one of my exams
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Let
$ x^3=5$
$ y^3=4$
$ 9\sqrt{\strut x-y}$
$ =(x^3+y^3)\sqrt{\strut x-y}$
$ =(x^2-xy+y^2)\sqrt{\strut(x-y)(x+y)^2}$
$ =(x^2-xy+y^2)\sqrt{\strut x^3+x^2y-xy^2-y^3}$
Now we can substitute the value already
$ =(\sqrt[3]{\strut 25}-\sqrt[3]{\strut 20}+\sqrt[3]{\strut 16})\sqrt{\strut 5+\sqrt[3]{\strut 100}-\sqrt[3]{\strut80}-4}$
$ =(\sqrt[3]{\strut 25}-\sqrt[3]{\strut 20}+\sqrt[3]{\strut 16})\sqrt{\strut \sqrt[3]{\strut 10}^2-2\sqrt[3]{\strut10}+1}$
$ =(\sqrt[3]{\strut 25}-\sqrt[3]{\strut 20}+\sqrt[3]{\strut 16})\sqrt{\strut (\sqrt[3]{\strut 10}-1)^2}$
$ =(\sqrt[3]{\strut 25}-\sqrt[3]{\strut 20}+\sqrt[3]{\strut 16})(\sqrt[3]{\strut 10}-1)$
$ =\sqrt[3]{\strut 250}-\sqrt[3]{\strut 200}+\sqrt[3]{\strut 160}-\sqrt[3]{\strut 25}+\sqrt[3]{\strut 20}-\sqrt[3]{\strut 16}$
$ =5\sqrt[3]{\strut 2}-2\sqrt[3]{\strut 25}+2\sqrt[3]{\strut 20}-\sqrt[3]{\strut 25}+\sqrt[3]{\strut 20}-2\sqrt[3]{\strut 2}$
$ =3\sqrt[3]{\strut 2}-3\sqrt[3]{\strut 25}+3\sqrt[3]{\strut 20}$
Hence
$ 3\sqrt{\strut \sqrt[3]{\strut 5}-\sqrt[3]{\strut 4}}$
$ =\sqrt[3]{\strut 2}-\sqrt[3]{\strut 25}+\sqrt[3]{\strut 20}$
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How can find this limit? $\lim_{x\to-∞}(\sqrt{x^2+6x}-\sqrt{x^2-2x})$ $$\lim_{x\to-∞}(\sqrt{x^2+6x}-\sqrt{x^2-2x})$$
plugging in infinity gives infinity - infinity, what kind of manipulation can I do to solve this?
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Note that $$ \sqrt{x^2+6x}-\sqrt{x^2-2x} = \frac{8x}{\sqrt{x^2+6x}+\sqrt{x^2-2x}}, $$so our answer is the limit of the above as $ x \to -\infty $. But note that $ \sqrt {x^2 + 6x} + \sqrt {x^2 - 2x} \sim \mathcal{O}(2x) $, so the answer is $\frac{8}{2}=\boxed{4}$.
|
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Why is $x(\sqrt{x^2+1} - x )$ approaching $1/2$ when $x$ becomes infinite? Why is $$\lim_{x\to \infty} x\left( (x^2+1)^{\frac{1}{2}} - x\right) = \frac{1}{2}?$$
What is the right way to simplify this?
My only idea is: $$x((x²+1)^{\frac{1}{2}} - x) > x(x^{2^{0.5}}) - x^2 = 0$$
But 0 is too imprecise.
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You already know that $\sqrt{x^2 + 1} \sim x$. The problem is that that approximation isn't good enough for computing this limit.
However, you've done exercises to do this exact thing: use a differential approximation and the value of $\sqrt{x^2}$ to estimate the value of $\sqrt{x^2 + 1}$. e.g. maybe you've been asked to estimate $\sqrt{81.4}$ using the knowledge of $\sqrt{81}$.
To put it back in familiar symbols, set $f(t) = \sqrt{x^2 + t}$. Then differential approximation tells us that
$$ f(1) \approx f(0) + f'(0) \cdot (1 - 0) $$
or that
$$ \sqrt{x^2 + 1} \approx x + \frac{1}{2 \sqrt{x^2}} = x + \frac{1}{2x} $$
and now we can see that the limit probably should be $1/2$. There is one last problem, though: we need to make sure there isn't too much error in the differential approximation! And unfortunately, the description of the error term for differential approximation isn't good enough for our purposes, since it tells us what happens as $t \to 0$, not as $x \to \infty$.
While there are some tricks for doing this, I will instead give a more systematic approach: use Taylor series with remainder. Here is one form:
$$ f(1) = f(0) + f'(0) \cdot (1-0) + \frac{1}{2} f''(a) \cdot (1-0)^2 $$
where $a$ is some unknown value in the interval $[0,1]$. So we have
$$ \sqrt{x^2 + 1} = \sqrt{x^2} + \frac{1}{2 \sqrt{x^2} } - \frac{1}{8 (x^2 + a)^{3/2}}
= x + \frac{1}{2 x } - \frac{1}{8 (x^2 + a)^{3/2}} $$
where $a \in [0,1]$ (and varies with $x$). Plugging this into our original limit now gives us something we can rigorously show converges to $1/2$.
It may help to use the fact this is a strictly increasing function of $a$ to arrange the above as
$$ x + \frac{1}{2 x } - \frac{1}{8x^3} \leq \sqrt{x^2 + 1} \leq x + \frac{1}{2 x } - \frac{1}{8(x^2 + 1)^{3/2}} $$
Interestingly, even though we needed a first-order estimate to solve the problem, we could have used a zeroth-order Taylor series with remainder, or equivalently the mean value theorem:
$$ f(1) = f(0) + f'(a) \cdot (1-0) = x + \frac{1}{2 \sqrt{x^2 + a}} $$
so that
$$x + \frac{1}{2\sqrt{x^2+1}} \leq \sqrt{x^2 + 1} \leq x + \frac{1}{2 x}$$
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How to solve following limit I've been struggeling a bit with the following limit:
$\lim\limits_{x \to 0} \frac{a- \sqrt{a^2 - x^2}}{x^2}$
The solution is:
If a < 0 then -$\infty$ .
If a > 0 then $\frac{1}{2a}$
But I don't know how to get there.
Thank you.
|
$$\begin{align*}
\lim_{x\to0}\frac{a-\sqrt{a^2-x^2}}{x^2}
&= \lim_{x\to0}\frac{a-\sqrt{a^2-x^2}}{x^2}\cdot\frac{a+\sqrt{a^2-x^2}}{a+\sqrt{a^2-x^2}}\\
&= \lim_{x\to0}\frac{x^2}{x^2\left(a+\sqrt{a^2-x^2}\right)}\\
&= \lim_{x\to0}\frac{1}{a+\sqrt{a^2-x^2}}\\
&= \frac{1}{a+\sqrt{a^2}}\\
\end{align*}$$
Now the tricky part is how you simplify $\sqrt{a^2}$ base on the sign of $a$.
|
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|
Simplifying $\sqrt {1+\sqrt 5}$ I considered simplifying $\sqrt {1+\sqrt 5}$.
So I started $(a+b \sqrt 5)^2 = 1 + \sqrt 5$.
This gave me $a\color{blue}{^2} + 5b^2 =1 , 2ab = 1$ so the result was
$ \sqrt{1+\sqrt 5} = \sqrt{1/2 - i} + \dfrac{\sqrt 5}{\sqrt{2 - 4i}}.$
I continued in the same spirit : $(a_2+b_2 i)^2 = 1/2 - i. $
$a_2^2 - b_2^2 = 1/2, 2a_2b_2 = -1.$ This gave me :
$ \sqrt{1+\sqrt 5} = \dfrac{\sqrt{1+\sqrt 5}}{2} - \frac{1}{4} (\sqrt 5 - 1)(\sqrt{1+\sqrt 5})\color{blue}i +\dfrac{\sqrt 5}{\sqrt{2 - 4i}}\tag1$
I was a bit surprised by the self-similarity.
I got the feeling of " periodic " or " fractal " , like running in circles.
It this normal to get this kind of self-similarity when the initial radical cannot be rewritten in simpler shorter form ( without complex numbers or nesting ) ?
Are they ways to compute or theorems to know about just how many steps it takes to get the self-similarity ?
Does this imply that the number of nontrivial non-nested expressions for a real radical is Always finite ?
edit
Perhaps this edit clarifies a bit what I meant.
Basicly the only identities and substitutions for identities we can have are :
$ \sqrt{1+\sqrt 5} = \sqrt{1/2 - i} + \sqrt{1/2 + i}$
$ \sqrt{1/2+ i} = \dfrac{\sqrt{1+\sqrt 5}}{2} - \frac{1}{4} (\sqrt 5 - 1)(\sqrt{1+\sqrt 5})i$
$ \sqrt{1/2- i} = \dfrac{\sqrt{1+\sqrt 5}}{2} + \frac{1}{4} (\sqrt 5 - 1)(\sqrt{1+\sqrt 5})i$
And no more than those 3. Or 2 if we do not count conjugates as seperate.
Now I suspect this mysterious number 2 is no coincidence ;
Conjecture
If a nested root form containing only sqrts , positive integers and additions (thus no divisions or substractions ) is given that cannot be simplified within the same ring then the amount of identities ( as in the example above , ignoring conjugates ) is at most the amount of sqrt's in the expression.
In this case 2 sqrt's => 2 identities.
edit
comment
A similar thing happens with $\sqrt{5+\sqrt 3}$. You end up with a $\sqrt{10+2\sqrt 22}$ term and when you go on you end up with $\sqrt{5+\sqrt 3}$ again.
Here $\sqrt{10+2\sqrt 22}$ takes on the " role " of $\sqrt{1/2 + i}$ but this examples demonstrates that the phenomenon can stay within the reals too.
These examples are fairly easy to check and in that sense they may appear trivial , but there must be more behind this.
The same thing happens for cuberoots for example , and that gives us a mimimum polynomial of degree 9 which is much less well understood then degree 4.
Similar things happen if we use Bring radicals.
Although intuitive and easy to check any particular example by basic algebra , group and ring theory ... The basic framework is unclear to me.
I do not think Galois theory alone explains it completely.
I gave a bounty but without succes.
|
(Too long for a comment.)
I think what he means is that since from $(1)$,
$$ \sqrt{1+\sqrt 5} = \dfrac{\sqrt{1+\sqrt 5}}{2} - \frac{1}{4} (\sqrt 5 - 1)(\sqrt{1+\sqrt 5}) i +\dfrac{\sqrt 5}{\sqrt{2 - 4i}}\tag1$$
Or,
$$ \dfrac{\sqrt{1+\sqrt 5}}{2} = -\frac{\sqrt 5-1}{4}\Bigr(\sqrt{1+\sqrt 5}\Bigl) i +\dfrac{\sqrt 5}{\sqrt{2 - 4i}}$$
So when he "... was surprised by the self-similarity", I assume it was because he saw that part of the RHS can be expressed in terms of the LHS,
$$ A = \color{blue}{\frac{1-\sqrt 5}{2} A i +\dfrac{\sqrt 5}{\sqrt{2 - 4i}}}$$
and we can iterate the "rule" for $A$ as,
$$ A = \frac{1-\sqrt 5}{2} \left(\color{blue}{\frac{1-\sqrt 5}{2} A i +\dfrac{\sqrt 5}{\sqrt{2 - 4i}}}\right) i +\dfrac{\sqrt 5}{\sqrt{2 - 4i}}$$
and again,
$$ A = \frac{1-\sqrt 5}{2} \left({\frac{1-\sqrt 5}{2} \left(\color{blue}{\frac{1-\sqrt 5}{2} A i +\dfrac{\sqrt 5}{\sqrt{2 - 4i}}}\right) i +\dfrac{\sqrt 5}{\sqrt{2 - 4i}}}\right) i +\dfrac{\sqrt 5}{\sqrt{2 - 4i}}$$
and so on to as many levels as one wishes, with $A = \frac{\sqrt{1+\sqrt 5}}{2}$, hence his "fractal".
|
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Calculate $\sqrt{\frac{1}{2}} \times \sqrt{\frac{1}{2} + \frac{1}{2}\sqrt{\frac{1}{2}}} \times \ldots $ $$
\sqrt{\frac{1}{2}} \times \sqrt{\frac{1}{2} + \frac{1}{2}\sqrt{\frac{1}{2}}} \times \sqrt{\frac{1}{2} + \frac{1}{2}\sqrt{\frac{1}{2}+ \frac{1}{2}\sqrt{\frac{1}{2}}}} \times\ldots$$
I already know a way to calculate it:
With $\cos{\frac{\pi}{4}} = \sqrt{\frac{1}{2}}$ and denote $\frac{\pi}{4} = x$. Observe that:
$$\sqrt{\frac{1}{2} + \frac{1}{2}\sqrt{\frac{1}{2}}} = \sqrt{\frac{1+\cos{x}}{2}} = \cos{\frac{x}{2}} \\
\sqrt{\frac{1}{2} + \frac{1}{2}\sqrt{\frac{1}{2}+ \frac{1}{2}\sqrt{\frac{1}{2}}}} = \cos\frac{x}{4}$$
Thus it becomes
\begin{align}
P(n) &= \cos{\frac{x}{2^n}}\cos{\frac{x}{2^{n-1}}} \cdots \cos{x} \\
&= \frac{2\sin{\frac{x}{2^{n-1}}}\cos{\frac{x}{2^{n-1}}}\cos{\frac{x}{2^{n-2}}} \cdots\cos{\frac{x}{2}}}{2\sin{\frac{x}{2^{n-1}}}}
\end{align}
Taking in to account that $2\sin{x}\cos{x} = \sin{2x}$, we have
\begin{align}
P(n) &= \lim_{n \to \infty} \frac{\sin{2x}}{2^n\sin{\frac{x}{2^{n-1}}}} \\
&= \lim_{n\rightarrow \infty} \frac{\sin{2x}}{2x} \\
&= \frac{2\sin\frac{\pi}{2}}{\pi} \\
&= \boxed{\frac{2}{\pi}}
\end{align}
Now, I'm looking for another solution, please comment on.
|
(Solution by OP in question: converted to a community wiki answer)
I already know a way to calculate it:
$$\cos{\frac{\pi}{4}} = \sqrt{\frac{1}{2}}, \frac{\pi}{4} = x$$
$$\sqrt{\frac{1}{2} + \frac{1}{2}\sqrt{\frac{1}{2}}} = \sqrt{\frac{1+\cos{x}}{2}} = \cos{\frac{x}{2}}$$
$$\sqrt{\frac{1}{2} + \frac{1}{2}\sqrt{\frac{1}{2}+ \frac{1}{2}\sqrt{\frac{1}{2}}}} = \cos\frac{x}{4}$$
Thus it becomes
$$P(n) = \cos{\frac{x}{2^n}}\cos{\frac{x}{2^{n-1}}} \cdots \cos{x}$$
$$P(n) = \frac{2\sin{\frac{x}{2^{n-1}}}\cos{\frac{x}{2^{n-1}}}\cos{\frac{x}{2^{n-2}}} \cdots\cos{\frac{x}{2}}}{2\sin{\frac{x}{2^{n-1}}}}$$
$$2\sin{x}\cos{x} = \sin{2x}$$
$$\lim_{n\rightarrow \infty} \frac{\sin{2x}}{2^n\sin{\frac{x}{2^{n-1}}}}$$
$$\lim_{n\rightarrow \infty} \frac{\sin{2x}}{2x}$$
$$\frac{2\sin\frac{\pi}{2}}{\pi} = \boxed{\frac{2}{\pi}}$$
|
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"url": "https://math.stackexchange.com/questions/1081519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "24",
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|
What is wrong with the sum of these two series? Could anyone help me to find the mistake in the following problem? Based on the formula of the sum of a geometric series:
\begin{equation}
1 + x + x^{2} + \cdots + x^{n} + \cdots = \frac{1}{1 - x}
\end{equation}
\begin{equation}
1 + \frac{1}{x} + \frac{1}{x^{2}} + \cdots + \frac{1}{x^{n}} + \cdots = \frac{1}{1 - 1/x} = \frac{x}{x-1}
\end{equation}
Adding both equations
\begin{equation}
2 + x + \frac{1}{x} + x^{2} + \frac{1}{x^{2}} + \cdots + x^{n} + \frac{1}{x^{n}} + \cdots = \frac{1}{1 - x} + \frac{x}{x-1} = \frac{1-x}{1-x} = 1
\end{equation}
So,
\begin{equation}
2 + x + \frac{1}{x} + x^{2} + \frac{1}{x^{2}} + \cdots + x^{n} + \frac{1}{x^{n}} + \cdots = 1
\end{equation}
And the left side is always bigger than $2$ for $x>0$.
What is wrong?? Thanks in advance
|
The domains of convergence of these two sequences don't coincide. One converges for $|x|>1$ and the other for $|x| < 1$. Therefore, the sum is meaningless.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Maximum of $f(x) = (45-2x)\cdot (24-2x)\cdot (2x)\;,$ Where $0How Can I Maximise $f(x) = (45-2x)\cdot (24-2x)\cdot (2x)\;,$ Where $0<x < 12$
Using Inequality
$\bf{My\; Try::}$ In $0<x<12\;,$ The value of $(45-2x)\;,(24-2x)\;,2x>0$
and we can write $\displaystyle f(x) = \frac{1}{2}\bigg[(45-2x)\cdot (24-2x)\cdot (4x)\bigg]$
So Applying $\bf{A.M\geq G.M\;,}$ We get
$\displaystyle \frac{(45-2x)+(24-2x)+4x}{3}\geq \bigg[(45-2x)\cdot (24-2x)\cdot (4x) \bigg]^{\frac{1}{3}}$
and equality hold when $(45-2x)= (24-2x)=(4x)\;,$ But this is wrong bcz no common
value of $x$ for which equality holds.
Help required, Thanks
|
We can also use AM-GM as follows:
$$f(x) = \frac{1}{2\cdot 5\cdot 7} [2(45-2x)]\cdot [5(24-2x)] \cdot [7(2x)]$$
$$\leq \frac{1}{2\cdot 5\cdot 7\cdot 27}\left[2(45-2x) + 5(24-2x) + 7(2x)\right]^3 = \frac{210^3}{2\cdot 5\cdot 7\cdot 27} = 70^2$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1082577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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|
How $\frac{1}{\sqrt{2}}$ can be equal to $\frac{\sqrt{2}}{2}$? How $\frac{1}{\sqrt{2}}$ can be equal to $\frac{\sqrt{2}}{2}$?
I got answer $\frac{1}{\sqrt{2}}$, but the real answer is $\frac{\sqrt{2}}{2}$. Anyway, calculator for both answers return same numbers.
|
One thing we tend to like to do when dealing with this stuff is have the denominator of fractionish things as a plain number, as much as possible. To do this, we'll multiply both top and bottom by something that will cancel out any radicals in the bottom.
$$\sqrt{\frac{1}{2}}=\frac{\sqrt{1}}{\sqrt{2}}=\frac{1}{\sqrt{2}}\cdot\frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}}{2}$$
This even works for more complicated stuff.
$$\frac{1}{\sqrt{5}+\sqrt{3}}=\frac{1}{\sqrt{5}+\sqrt{3}}\cdot\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}-\sqrt{3}}=\frac{\sqrt{5}-\sqrt{3}}{5-3}=\frac{\sqrt{5}-\sqrt{3}}{2}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1082664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
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|
Find Min Values Of $P=\frac{1}{(1+x)^2}+\frac{1}{(1+y)^2}+\frac{1}{(1+z)^2}+\frac{4}{(1+x)(1+y)(1+z)}$ Given $x,y,z>0$ and $y+z=x(y^2+z^2)$
Find Min Values Of
$P=\frac{1}{(1+x)^2}+\frac{1}{(1+y)^2}+\frac{1}{(1+z)^2}+\frac{4}{(1+x)(1+y)(1+z)}$
Could Someone give me an idea ?
|
by the Lagrange Multiplier Method we get
$$P\geq \frac{91}{108}$$ the equal sign holds if $$x=\frac{1}{5},y=z=5$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1082914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
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|
Polynomial annihilator method $y''+8y=5x+2e^{-x}$ $y''+8y=5x+2e^{-x}$
It is asked to solve the equation by this method.
My result was $c_1\sin(\sqrt8x)+c_2\cos(\sqrt8x)+5x+\frac{2}{9}e^{-x}$
Then it is aked a particular solution so that $y(0)=0$ and $y'(0)=-{1/15}$
I arrived to $c_2=-\frac{2}{9}$ and to $c_1=-\frac{218}{45}$
By the last result I think something might be wrong.
Thanks
|
We are given
$$\tag 1 y'' + 8y = 5x + 2e^{-x}, y(0)=0, y'(0)=-\dfrac 1{15}$$
We are asked to use the Annihilator Method, but here are some clearer notes that I will follow.
For the homogeneous part of $(1)$, we have:
$$(D^2+8)y = 5x + 2e^{-x} \implies D^2 + 8 = 0 \implies D = \pm 2 \sqrt{2}~i$$
This gives us a homogeneous solution of:
$$y_h(x) = c_1 \cos(2 \sqrt{2} x) + c_2 \sin(2 \sqrt{2} x)$$
For the Right-Hand-Side (RHS) of $(1)$, we have the annihilator of $D^2(D+1)$, see notes if that is not clear. Applying this to both sides of the original differential
equation we get:
$$D^2(D + 1)(D^2 + 8) y = 0 \implies D = \pm 2 \sqrt{2}~i, -1, 0, 0$$
The general solution to this homogeneous equation is:
$$y(x) = c_1 \cos(2 \sqrt{2} x) + c_2 \sin(2 \sqrt{2} x) + c_3 e^{-x} + c_4 + c_5 x$$
We eliminate the original homogeneous solution from this solution and that leaves our particular solution as:
$$y_p(x) = c_3 e^{-x} + c_4 + c_5 x$$
We now want to substitute this into $(1)$ to determine the coefficients and have:
$$y_p(x) = c_3 e^{-x} + c_4 + c_5 x, ~~y'_p(x) = -c_3 e^{-x} + c_5, ~~y''_p(x) = c_3e^{-x}$$
Substituting these into $(1)$ and equating terms yields:
$$c_3 e^{-x} + 8(c_3 e^{-x} + c_4 + c_5 x) = 9 c_3 e^{-x} + 8c_4 + 8 c_5 x = 2 e^{-x} + 5x$$
This gives $c_3 = \dfrac 29, c_4 = 0, c_5 = \dfrac 58$.
Thus our solution is:
$$y(x) = c_1 \cos(2 \sqrt{2} x) + c_2 \sin(2 \sqrt{2} x) + \dfrac 29 e^{-x} + \dfrac 58 x$$
Now, we are given the two initial conditions in $(1)$ to solve for $c_1$ and $c_2$.
From $y(0) = 0$, we have:
$$y(0) = c_1 + \dfrac 29 = 0 \implies c_1 = -\dfrac 29$$
From $y'(0)=-\dfrac 1{15}$, we need the derivative of the solution, which is:
$$y'(x) = -2 \sqrt{2} c_1 \sin(2 \sqrt{2} x) + 2 \sqrt{2} c_2 \cos(2 \sqrt{2} x) -\dfrac 29 e^{-x} + \dfrac 58$$
So, we have:
$$y'(0) = 2 \sqrt{2} c_2 -\dfrac 29 + \dfrac 58 = -\dfrac 1{15} \implies 2 \sqrt{2} c_2 = - \dfrac {169}{360} \implies c_2 = -\dfrac {169}{720 \sqrt{2}}$$
Thus, our final solution is:
$$y(x) = -\dfrac 29 \cos(2 \sqrt{2} x) -\dfrac {169}{720 \sqrt{2}} \sin(2 \sqrt{2} x) + \dfrac 29 e^{-x} + \dfrac 58 x$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Double integral $ \iint \limits_D \frac{y}{x^2+(y+1)^2}dxdy$, $D$=$\{(x,y): x^2+y^2 \le1 , y\ge0\}$
Solve $$ \iint \limits_D \frac{y}{x^2+(y+1)^2}dxdy \ \ \ \ . . . \ (*)$$
where $D$=$\{$$(x,y): x^2+y^2 \le1 , y\ge0 $$\}$
$$
$$
Here is my attempt.
$$\begin{align}
&(1).\ \ \ (*)=\int_{-1}^1 \int_{0}^{\sqrt{1-x^2}}\frac{y}{x^2+(y+1)^2}dydx \\
&(2).\ \ \ (*)= \int_{0}^1 \int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}\frac{y}{x^2+(y+1)^2}dxdy \\
&(3). \ \ \int\frac{y+1}{x^2+(y+1)^2}dx = \arctan\left(\frac{x}{y+1}\right) + C \\
&(4). \ \ \ (*)=\int_{0}^{\pi} \int_{0}^{1}\frac{r^2sin\theta}{r^2+2rsin\theta+1}drd\theta \\\\
\end{align}$$
I used $(1)$, $(4)$ and $(2)$ with $(3)$,
but didn't solve yet.
$$$$
Did I make a mistake?
Could you give me some advice, please?
How can I solve this integral...
Thank you for your attention to this matter.
$$$$
P.S.
Here is result of wolframalpha
$$$$
$$ $$
Additionally... I did like this.. maybe useless :-(
$$\begin{align}
(*)
& = \int_{0}^1 \int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}\frac{y}{x^2+(y+1)^2}dxdy \\\\
&=\int_{0}^1 \int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}\frac{y+1}{x^2+(y+1)^2}dxdy + \int_{0}^1 \int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}\frac{1}{x^2+(y+1)^2}dxdy \\\\
&=\int_{0}^1 \left(\arctan\left(\frac{\sqrt{1-y^2}}{y+1}\right) - \arctan\left(\frac{-\sqrt{1-y^2}}{y+1}\right)\right)dy \\
& \ \ \ \ + \int_{0}^1 \int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}\frac{1}{x^2+(y+1)^2}dxdy \\\\
&=\int_{0}^1 \left(\arctan\left(\sqrt\frac{1-y}{1+y} \ \right) - \arctan\left(-\sqrt\frac{1-y}{1+y} \ \right)\right)dy \\
& \ \ \ \ +\int_{0}^1 \int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}\frac{1}{x^2+(y+1)^2}dxdy \\\\
&= terrible?! \\
\end{align}$$
$$
$$
$$ $$
---------------------------------------------------------------------------
This picture is for asking to Christian Blatter
(I am really sorry, if I bother you guys for this picture.)
|
The integrand function $\frac{y}{x^2+(y+1)^2}$ suggest to put
$$\left\{
\begin{align}
x&=r\cos\theta\\
y+1&=r\sin\theta
\end{align}\right.
$$
so that $x^2+(y+1)^2=r^2$ and the Jacobian is $r$.
From $y\ge 0$ we have $y=r\sin\theta-1\ge 0$ that is $r\ge\frac{1}{\sin\theta}$ and from $x^2+y^2\le 1$ we have $$x^2+y^2=r^2\cos^2\theta+(r\sin\theta-1)^2=r^2-2r\sin\theta+1\le 1$$ and then $r(r-2\sin\theta)\le0$ so that $r\le 2\sin\theta$.
So we have
$$\boxed{
r_{\min}=\frac{1}{\sin\theta}\le r\le 2\sin\theta=r_{\max}}
$$
For $y=0$ (i.e. $r\sin\theta =1$) we have $-1\le x\le 1$, that is $-1\le r\cos\theta\le 1$ and then $-1\le\tan\theta\le 1$; thus
$$\boxed{
\theta_{\min}=\frac{\pi}{4}\le \theta\le \frac{3\pi}{4}=\theta_{\max}}
$$
or $\frac{-\pi}{4}\le \theta\le \frac{+\pi}{4}$ if you prefer.
The figure help to show all we have done.
So the integrand in polar coordinates becomes $f(r,\theta)=\frac{r\sin\theta-1}{r^2}$ and the integral becomes
$$
\mathcal{I}=\int_{\theta_{\min}}^{\theta_{\max}}\int_{r_{\min}}^{r_{\max}}
f(r,\theta)r\,\mathrm d r\,\mathrm d\theta=
\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}}\int_{\frac{1}{\sin\theta}}^{2\sin\theta}\left(\sin\theta-\frac{1}{r}\right) \mathrm d r\,\mathrm d\theta
$$
The integral in $r$ is easy to evaluate
$$\begin{align}
\int_{\frac{1}{\sin\theta}}^{2\sin\theta}\left(\sin\theta-\frac{1}{r}\right) \mathrm d r
&=
\left[\sin\theta\, r-\log r\right]_{\frac{1}{\sin\theta}}^{2\sin\theta}\\
&=\sin\theta\left[2\sin\theta-\tfrac{1}{\sin\theta}\right]-\left[\log(2\sin\theta)-\log\left(\tfrac{1}{\sin\theta}\right)\right]\\
&=-\cos(2\theta)-\log\left(2\sin^2\theta\right)
\end{align}
$$
Then the integral in $\theta$ is
$$\begin{align}
\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \left[-\cos(2\theta)-\log\left(2\sin^2\theta\right)\right]\mathrm d \theta
&=
\left[-\frac{1}{2}\sin(2\theta)\, \right]_{\frac{\pi}{4}}^{\frac{3\pi}{4}}-\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \log\left(2\sin^2\theta\right)\mathrm d \theta=1+J
\end{align}
$$
where
$$
J=-\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \log\left(2\right)\mathrm d \theta-\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \log\left(\sin^2\theta\right)\mathrm d \theta=
-\frac{\pi}{2}\log 2-2C+\pi\log 2=\frac{\pi}{2}\log 2-2C
$$
observig that
$$\begin{align}
-\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \log\left(\sin^2\theta\right)\mathrm d \theta &=
-\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \log\left(\cos^2\theta\right)\mathrm d \theta=-2\int_{0}^{\frac{\pi}{4}} \log\left(\cos^2\theta\right)\mathrm d \theta\\
&=-4\int_{0}^{\frac{\pi}{4}} \log\left(\cos\theta\right)\mathrm d \theta=-4\left(\frac{C}{2}-\frac{\pi}{4}\log 2\right)\\
&=-2C+\pi\log 2
\end{align}
$$
where $C$ is the Catalan's constant (see for exaple here).
Finally we have
$$\large\color{blue}{
\mathcal I=1+\frac{\pi}{2}\log 2-2C}
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How prove $\bigl(\frac{\sin x}{ x}\bigr)^{2} + \frac{\tan x }{ x} >2$ for $0 < x < \frac{\pi}{2}$ How prove $\left(\frac{\sin x}{ x}\right)^{2} + \frac{\tan x }{ x} >2$ for $0 < x < \frac{\pi}{2}$.
Can this be proved with simple way?
|
$$\left(\frac{\sin x}{x}\right)^2 + \left(\frac{\tan x}{x}\right) = \\ \left(\frac{x - \frac{x^3}{6} + \frac{x^5}{120} + \ldots}{x}\right)^2 + \left(\frac{x + \frac{x^3}{3} + \frac{2x^5}{15} + \ldots}{x}\right) = \\ \left(1 - \frac{x^2}{6} + \frac{x^4}{120} - \ldots\right)^2 + \left(1 + \frac{x^2}{3} + \frac{2x^4}{15} + \ldots\right) \simeq \\ \left(1 + 2 \left(-\frac{x^2}{6} + \frac{x^4}{120} - \ldots \right)\right) + 1 + \frac{x^2}{3} + \frac{2x^4}{15} + \ldots\\2 + x^2 \left(\frac{ - 2}{6} + \frac{1}{3}\right) + x^4 \left(\frac{2}{120} + \frac{2}{15} \right) + \ldots = \\2 + x^4 \left(\frac{9}{60} \right) + \ldots > 2.$$
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
What is the appropriate substitution for this equation? I need to solve this equation:
$x^\frac{1}{3}+x^\frac{1}{7}=2(x^5)^\frac{1}{21}.$
I can't find the substitution which reduces it to a system or equation without roots.
|
$$\text{The equation can be rewritten as follows:}$$
$$x^{\frac{7}{21}} + x^{\frac{3}{21}} = 2(x^{\frac{5}{21}})$$
$$$$
$$\text{Now, let } y = x^{\frac{1}{21}}:$$
$$y^7 + y^3 = 2y^5 \implies y^7 - 2y^5 + y^3 = 0$$
$$$$
$$\text{If } y = 0, \text{ then } x = 0, \text{ else we have: }$$
$$y^4 - 2y^2 + 1 = 0 \implies (y^2 - 1)^2 = 0 \implies y^2 = 1$$
$$$$
$$\text{Hence, } y \in \{-1, 0, 1\}.$$
$$\text{And thus, } x \in \{-1, 0, 1\} \text{ since } x = y^{21}.$$
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
$y=\frac{1}{x}$, show that $\frac{dy}{dt}=\frac{-1}{x^2}\frac{dx}{dt}$ and find $\frac{d^2y}{dt^2}$ If $y=\frac{1}{x}$, where $x$ is a function of $t$ show that $\frac{dy}{dt}=\frac{-1}{x^2}\frac{dx}{dt}$ and find an expression for $\frac{d^2y}{dt^2}$.
For $\frac{d^2y}{dt^2}$ I keep getting
$$\frac{d^2y}{dt^2}=\frac{-1}{x^2}\frac{d^2x}{dt^2}+\frac{2}{x^3}\frac{dx}{dt}$$
but the correct answer is apparently
$$\frac{d^2y}{dt^2}=\frac{-1}{x^2}\frac{d^2x}{dt^2}+\frac{2}{x^3}\left({\frac{dx}{dt}}\right)^2$$
|
$$
\dfrac{d}{dt}g(x)\dot{x}= \dfrac{dg}{dt}\dot{x} + g(x)\ddot{x}
$$
Now let's have $g(x) = -1/x^2$
We have
$$
\dfrac{dg}{dt} = \dfrac{dg}{dx}\dot{x} = \frac{2}{x^3}\dot{x}
$$
So hopefully it is clear to see that you have compute the derivative of $-1/x^2$ wrong?
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove convergence of series Let $$\displaystyle a_n=\sum_{n=2}^{\infty} \frac{(-1)^n}{n^{\frac{1}{3}}+(-1)^{\frac{n(n+1)}{2}}}$$
so I divide it into four series $4k, 4k+1, 4k+2, 4k+3$ and I pair for instance $4k, 4k+1$ and $4k+2, 4k+3$ and prove that these two series is convergent and conclude that since
both series are convergent so the sum of it is also convergent but I'm not sure if it's legall
second series I'm not sure is like this $\displaystyle b_n=\frac{1}{1}+\frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+\frac{1}{9}+\frac{1}{11}-\frac{1}{6}+...$
here I also split it into smaller parts namely $\displaystyle \sum_{n=1}^{\infty}\frac{1}{4n-3}+\sum_{n=1}^{\infty}\frac{1}{4n-1}-\sum_{n=1}^{\infty}\frac{1}{2n}$ and I prove that $\displaystyle \frac{1}{4n-3}+\frac{1}{4n-1}-\frac{1}{2n}<\frac{1}{n^2} \cdot c $ for some const c so it implies that origin series is convergent or not ?
|
Hint. Here is an approach.
Recall that, for $x$ near $0$, you have by the Taylor series expansion or simply by the finite sum of a geometric sequence:
$$
\frac{1}{1+x}=1-x+x^2+\mathcal{O}(x^3),
$$
You may then write, as $n$ is great:
$$
\frac{(-1)^n}{n^{1/3}+(-1)^{\frac{n(n+1)}{2}}}=\frac{(-1)^n}{n^{1/3}\left(1+\frac{(-1)^{\frac{n(n+1)}{2}}}{n^{1/3}}\right)}=\frac{(-1)^n}{n^{1/3}}-\frac{(-1)^{\frac{n(n+3)}{2}}}{n^{2/3}}+\mathcal{O}\left(\frac{1}{n^{4/3}}\right)
$$ The series $ \displaystyle
\sum\frac{(-1)^n}{n^{1/3}}$
is convergent by the alternating series test, the series $\displaystyle
\sum\frac{1}{n^{4/3}}$ is convergent $(4/3>1)$. Since
$$
\left|\sum_{k=2}^{2N}(-1)^{n(n+3)/2}\right|\leq 2, \qquad \left|\sum_{k=2}^{2N+1}(-1)^{n(n+3)/2}\right|\leq 2,\quad N=1,2,3\ldots,
$$ then the series $ \displaystyle
\sum\frac{(-1)^{\frac{n(n+3)}{2}}}{n^{2/3}}$
is convergent by the Dirichlet series test.
|
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|
Formalize a proof without words of the identity $(1 + 2 + \cdots + n)^2 = 1^3 + 2^3 + \cdots + n^3$ This website gives the following proof without words for the identity $(1 + 2 + \cdots + n)^2 = 1^3 + 2^3 + \cdots + n^3$.
I find it interesting but have trouble seeing the proof behind it. Could anyone could give me a formal proof based on this animation?
|
This gif is a nice visualization of the following algebraic manipulations (color to correspond to the colors in the gif):
\begin{align*}
(1 + 2 + 3 + \cdots + n)^2
&= \sum_{i=1}^n \sum_{j=1}^n ij \\
&= \color{green}{\sum_{1 \le i < j \le n} ij} + \color{red}{\sum_{1 \le j < i \le n} ji} + \color{blue}{\sum_{1 \le j \le n} j^2} \\
&= \color{green}{\sum_{1 \le i < j \le n} ij} + \color{red}{\sum_{1 \le i < j \le n} ji} + \color{blue}{\sum_{1 \le j \le n} j^2} \\
&= \sum_{j=1}^n j \left[ \color{green}{\sum_{i=1}^{j-1} i} + \color{red}{\sum_{i=1}^{j-1} i} + \color{blue}{j} \right] \\
&= \sum_{j=1}^n j \left[ \sum_{i=1}^{j-1} \big(\color{green}{(i)} + \color{red}{(j-i)}\big) + \color{blue}{j} \right] \\
&= \sum_{j=1}^n j \left[ \sum_{i=1}^j j\right] \\
&= \sum_{j=1}^n j^3.
\end{align*}
|
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|
If $0 \leq a, b, c \leq 1$, the inequality $| a - b + c| \cdot |(a - c)^2 + b^2| \leq 1$ holds Please, help me to prove the inequality or find a counter-example
For $0 \leq a, b, c \leq 1$, prove that
$|a - b + c| \cdot |(a - c)^2 + b^2| \leq 1$.
|
*
*$|a-c|\le 1 \implies(a-c)^2\le |a-c|$
*$b\le 1 \implies b^2\le b$
$\therefore (a-c)^2+b^2\le |a-c|+b=a-c+b=a+(b-c)$$\hspace{10mm}$(if $a\ge c$)
Now
\begin{align}
| a - b + c| \cdot |(a - c)^2 + b^2 | & = |[a - b + c] \cdot [(a - c)^2 + b^2] | \\[6pt]
& \le |[a - b + c] \cdot [a+(b-c)]| \\[6pt]
& = |[a - (b - c)] \cdot [a+(b-c)]| \\[6pt]
& = | a^2 - (b - c)^2| \\[6pt]
&\le 1 \text{ (because $a^2 \le 1;(b - c)^2\le1$ )}
\end{align}
Similarly,
if $c\ge a$ we get $$ (a-c)^2+b^2\le |a-c|+b=c-a+b=c+(b-a)$$
Now
\begin{align}
| a - b + c| \cdot |(a - c)^2 + b^2 | & = |[a - b + c] \cdot [(a - c)^2 + b^2] | \\[6pt]
& \le |[a - b + c] \cdot [c+(b-a)]| \\[6pt]
& = |[c - (b - a)] \cdot [c+(b-a)]| \\[6pt]
& = | c^2 - (b - a)^2| \\[6pt]
&\le 1 \text{ (because $c^2 \le 1;(b - a)^2\le1$ )}
\end{align}
$\therefore$ In any case
$$|a - b + c| \cdot |(a - c)^2 + b^2| \le1$$
|
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|
Evaluate $\int\frac{\cos^2 x}{1-\sin x }dx$ $\int\frac{\cos^2 x}{1-\sin x} dx $ can someone explain me how to solve this one and please show your complete solution? So am I supposed to make the numerator $1+sinx$? but I think that doesn't help. Should I do long division?
|
We know that $\displaystyle \cos^2 x = 1 - \sin^2 x = (1-\sin x)(1+ \sin x)$
Hence,
$$
\require{cancel}
\begin{align}
\int\frac{\cos^2x}{1-\sin x}\mathrm dx &= \int \frac{\cancel{(1-\sin x)}(1+\sin x)}{\cancel{1- \sin x}}\mathrm dx\\
&= \int \mathrm dx + \int \sin x\\
&= x - \cos x + \color{gray}{\mathcal C}
\end{align}$$
Aliter:: The same method in another approach:
$$\require{cancel} \frac{\cos^2 x}{1 - \sin x}\cdot \frac{1 + \sin x}{1 + \sin x} = \frac{\cancel{\cos ^2 x}\cdot (1 + \sin x)}{\cancel{\cos^2 x}} = 1 + \sin x$$
|
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|
$\frac{(2n)!}{4^n n!^2} = \frac{(2n-1)!!}{(2n)!!}=\prod_{k=1}^{n}\bigl(1-\frac{1}{2k}\bigr)$
i cant see why we have :
*
*$$\frac{(2n)!}{4^n n!^2} = \frac{(2n-1)!!}{(2n)!!}$$
*$$\dfrac{(2n-1)!!}{(2n)!!} =\prod_{k=1}^{n}\left(1-\dfrac{1}{2k}\right),$$
Even i see the notion of Double factorial
this question is related to that one : Behaviour of the sequence $u_n = \frac{\sqrt{n}}{4^n}\binom{2n}{n}$
*
*For $\frac{(2n)!}{4^n n!^2} = \frac{(2n-1)!!}{(2n)!!}$
$\dfrac{(2n)!}{4^n n!^2}=\dfrac{(2n)!}{2^{2n} n!^2}=\dfrac{(2n)\times (2n-1)!}{2^{2n} (n\times (n-2)!)^2}$
*
*For $\dfrac{(2n-1)!!}{(2n)!!} =\prod_{k=1}^{n}\left(1-\dfrac{1}{2k}\right),$
note that $n!! = \prod_{i=0}^k (n-2i) = n (n-2) (n-4) \cdots$
$\dfrac{(2n-1)!!}{(2n)!!}=\dfrac{\prod_{i=0}^k (2n-1-2i)}{\prod_{i=0}^k (2n-2i)}$
|
$\dfrac{(2n)!}{4^nn!^2}=\dfrac{(2n)!!(2n-1)!!}{(2^nn!)^2}=\dfrac{(2n)!!(2n-1)!!}{((2n)!!)^2}=\dfrac{(2n-1)!!}{(2n)!!}$
$\prod_{k=1}^n(1-\frac{1}{2k})=\prod_{k=1}^n\frac{2k-1}{2k}=\dfrac{\prod_{k=1}^n(2k-1)}{\prod_{k=1}^n(2k)}=\frac{(2n-1)!!}{(2n)!!}$
|
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|
Arithmetic pattern $1 + 2 = 3$, $4 + 5 + 6 = 7 + 8$, and so on I am noticing this pattern:
\begin{align}
1+2&=3\\
4+5+6&=7+8\\
9+10+11+12&=13+14+15 \\
16+17+18+19+20&=21+22+23+24 \\
&\vdots
\end{align}
Is there a general formula that expresses a pattern here, such as for the $n$th term, or the closed form of the entire summation? I think the $n$th term starts with $$n^2+(n^2+1)+\cdots=\cdots+[(n+1)^2-1].$$
I am also presuming once this formula is discovered, we can prove it by induction for any $n$.
|
Both sides are $n (n+1)(2n+1)/2$ which also happens to be
$3 (1^2 + 2^2 + \ldots + n^2)$.
|
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|
How prove this limits $\lim_{n\to\infty}\frac{v_{5}(1^1\cdot 2^2\cdot 3^3\cdot 4^4\cdots\cdot n^n)}{n^2}=\frac{1}{8}$ Interesting Question:
Let denote by $v_{p}(a)$ the exponent of the prime number $p$ in the prime factorization of $a$,
show that
$$\lim_{n\to\infty}\dfrac{v_{5}(1^1\cdot 2^2\cdot 3^3\cdot 4^4\cdots\cdot n^n)}{n^2}=\dfrac{1}{8}$$
My some idea: since
$$1^1\cdot 2^2\cdot 3^3\cdot 4^4\cdots\cdot n^n=\dfrac{(n!)^n}{1!\cdot 2!\cdot 3!\cdots (n-1)!}$$
and it is well know
$$v_{5}(n!)=\lfloor \dfrac{n}{5}\rfloor+\lfloor\dfrac{n}{5^2}\rfloor+\lfloor\dfrac{n}{5^3}\rfloor+\cdots+\lfloor\dfrac{n}{5^k}\rfloor+\cdots=\sum_{i=1}^{\infty}\lfloor\dfrac{n}{5^k}\rfloor$$
so
$$v_{5}(1^1\cdot 2^2\cdot 3^3\cdot 4^4\cdots\cdot n^n)=n\sum_{i=1}^{\infty}\lfloor\dfrac{n}{5^k}\rfloor-\sum_{i=1}^{n-1}\sum_{k=1}^{\infty}\left(\lfloor\dfrac{i}{5^k}\rfloor\right)$$
then I can't it.
Thank you
|
I think it is better to count directly.
We have $5+10+15+...+5[n/5]=5\frac{[n/5]([n/5]+1)}{2}$ counts one factor $5$ from each multiple of $5$ raised to the power the number has.
Then $25+50+75+...+25[n/25]=25\frac{[n/25]([n/25]+1)}{2}$ counts one factor $5$ from each multiple of $25$ raised to the power it appears. Notice the other factor $5$ it provides was already been counted in the previous sum.
And so on, the next count is $125+250+...+125[n/125]=125\frac{[n/125]([n/125]+1)}{2}$.
We sum now all these sums
$$5\frac{[n/5]([n/5]+1)}{2}+25\frac{[n/25]([n/25]+1)}{2}+125\frac{[n/125]([n/125]+1)}{2}+...$$
It is enough to compute the limit for $n$ a power of $5$, $n=5^N$. The sum becomes
$$5\frac{5^{N-1}(5^{N-1}+1)}{2}+25\frac{5^{N-2}(5^{N-2}+1)}{2}+...+5^N\frac{1(1+1)}{2}\\=\frac{(5^{2N-1}+5^N)+(5^{2N-2}+5^{N})+...+(5^N+5^N)}{2}\\=\frac{5^N\frac{5^N-1}{5-1}+N5^N}{2}\\=\frac{5^{2N}-5^N+4N5^N}{8}$$
We must divide this by $5^{2N}$ and take limit, which is $1/8$. For $n$ moving along powers of 5 plus a bit the work is similar.
|
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|
Calculating $\sum_{n=1}^\infty\frac{1}{(n-1)!(n+1)}$ I want to calculate the sum:$$\sum_{n=1}^\infty\frac{1}{(n-1)!(n+1)}=$$
$$\sum_{n=1}^\infty\frac{n}{(n+1)!}=$$
$$\sum_{n=1}^\infty\frac{n+1-1}{(n+1)!}=$$
$$\sum_{n=1}^\infty\frac{n+1}{(n+1)!}-\sum_{n=1}^\infty\frac{1}{(n+1)!}=$$
$$\sum_{n=1}^\infty\frac{1}{n!}-\sum_{n=1}^\infty\frac{1}{(n+1)!}.$$
I know that $$\sum_{n=0}^\infty\frac{1}{n!}=e$$
so $$\sum_{n=1}^\infty\frac{1}{n!}=\sum_{n=0}^\infty\frac{1}{n!}-1=e-1$$
But, what can I do for $$\sum_{n=1}^\infty\frac{1}{(n+1)!}$$ ?
Am I allowed to start a sum for $n=-1$ ? How can I bring to a something similar to $$\sum_{n=0}^\infty\frac{1}{n!}$$?
|
$$f(x) = x e^{x} = \sum_{n=1}^{\infty} \frac{x^{n}}{(n-1)!} $$
Integrate to get
$$\int_0^x dt \, t e^t = \sum_{n=1}^{\infty} \frac{x^{n+1}}{(n+1)(n-1)!} $$
Integrate by parts...
$$\int_0^x dt \, t e^t = x e^x - \int_0^x dt \, e^t = (x-1) e^x +1$$
Plug in $x=1$ on both sides to get
$$\sum_{n=1}^{\infty} \frac{1}{(n+1)(n-1)!} = 1$$
|
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|
Closed form solution for $\sum_{n=1}^\infty\frac{1}{1+\frac{n^2}{1+\frac{1}{\stackrel{\ddots}{1+\frac{1}{1+n^2}}}}}$. Let
$$
\text{S}_k = \sum_{n=1}^\infty\cfrac{1}{1+\cfrac{n^2}{1+\cfrac{1}{\ddots1+\cfrac{1}{1+n^2}}}},\quad\text{$k$ rows in the continued fraction}
$$
So for example, the terms of the sum $\text{S}_6$ are
$$
\cfrac{1}{1+\cfrac{n^2}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+n^2}}}}}}
$$
Using a symbolic computation software (Mathematica), I got the following interesting results:
$$
\begin{align}
\text{S}_4 &= \frac{\pi}{4}\left(\coth(\pi)+\sqrt{3}\coth(\sqrt{3}\pi)\right)-\frac{1}{2}\\
\text{S}_6 &= \frac{\pi}{4}\left(\sqrt{2}\coth\left(\sqrt{2}\pi\right)+\sqrt{\frac{4}{3}}\coth\left(\sqrt{\frac{4}{3}}\pi\right)\right)-\frac{1}{2}\\
\text{S}_8 &= \frac{\pi}{4}\left(\sqrt{\frac{3}{2}}\coth\left(\sqrt{\frac{3}{2}}\pi\right)+\sqrt{\frac{7}{4}}\coth\left(\sqrt{\frac{7}{4}}\pi\right)\right)-\frac{1}{2}\\
\text{S}_{10} &= \frac{\pi}{4}\left(\sqrt{\frac{5}{3}}\coth\left(\sqrt{\frac{5}{3}}\pi\right)+\sqrt{\frac{11}{7}}\coth\left(\sqrt{\frac{11}{7}}\pi\right)\right)-\frac{1}{2}\\
\text{S}_{12} &= \frac{\pi}{4}\left(\sqrt{\frac{8}{5}}\coth\left(\sqrt{\frac{8}{5}}\pi\right)+\sqrt{\frac{18}{11}}\coth\left(\sqrt{\frac{11}{7}}\pi\right)\right)-\frac{1}{2}\\
\text{S}_{14} &= \frac{\pi}{4}\left(\sqrt{\frac{13}{8}}\coth\left(\sqrt{\frac{13}{8}}\pi\right)+\sqrt{\frac{29}{18}}\coth\left(\sqrt{\frac{29}{18}}\pi\right)\right)-\frac{1}{2}.\\
\end{align}
$$
The numbers appearing at the first $\coth$ term are easy to guess: they are the famous Fibonacci numbers.
The numbers at the second $\coth$ term can also be guessed: they appear to be the Lucas numbers. Those are constructed like the Fibonacci numbers but starting with $2,1$ instead of $0,1$.
Hence:
Conjecture: $$\text{S}_{2k}=\frac{\pi}{4}\left(\sqrt{\frac{F_k}{F_{k-1}}}\coth\left(\sqrt{\frac{F_k}{F_{k-1}}}\pi\right)+\sqrt{\frac{L_k}{L_{k-1}}}\coth\left(\sqrt{\frac{L_k}{L_{k-1}}}\pi\right)\right)-\frac{1}{2}$$
I have verified this conjecture for many $k$'s and it always work out perfectly. To me this is quite amazing, but I am not able to verify the conjecture. Can anyone prove it?
Moreover, if true the conjecture implies that
$$\lim_{k\to\infty}\text{S}_{2k}=\frac{\sqrt{\varphi}\pi\coth\left(\sqrt{\varphi}\pi\right)-1}{2}$$
which is also very nice ($\pi$ and $\varphi$ don't meet very often).
|
This is not a full answer but a partial result.
You can prove by induction that the described fraction of yours with $k$ horizontal lines is equal to $\frac{F_{k-1}n^2+F_{k}}{F_{k-2}n^4+2F_{k-1}n^2+F_k}$ with $k\ge3$. Perhaps with partial fractioning, you can compute the series using the well known result:
$$
\sum_{n=1}^{\infty}\frac{1}{n^2+z^2}=\frac{\pi z\coth(\pi z)-1}{2z^2}
$$
As a partial result you can compute the limit directly:
$$
\lim_{k\to\infty}\frac{F_{k-1}n^2+F_{k}}{F_{k-2}n^4+2F_{k-1}n^2+F_k}=\lim_{k\to\infty}\frac{F_{k-1}}{F_{k-2}}\frac{n^2+\frac{F_{k}}{F_{k-1}}}{n^4+2\frac{F_{k-1}}{F_{k-2}}n^2+\frac{F_k}{F_{k-2}}}=\varphi\frac{n^2+\varphi}{n^4+2\varphi n^2+\varphi^2}=\frac{\varphi}{n^2+\varphi}
$$
And therefore:
$$
\lim_{k\to\infty}\sum_{n=1}^\infty\frac{1}{1+\frac{n^2}{1+\frac{1}{\stackrel{\ddots}{1+\frac{1}{1+n^2}}}}}=\sum_{n=1}^{\infty}\frac{\varphi}{n^2+\varphi}=\varphi\frac{\pi \sqrt{\varphi}\coth(\pi \sqrt{\varphi})-1}{2\varphi}=\frac{\pi \sqrt{\varphi}\coth(\pi \sqrt{\varphi})-1}{2}
$$
I hope this is helpful.
|
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|
How is the integral $\int \frac{1}{ae^{mx}+be^{-mx}} \, dx$ solved? I am trying to determine how the following integral was solved, or at least the name of article deriving this solution:
$$
\int \frac{1}{ae^{mx}+be^{-mx}} \, dx
= \frac{1}{m\sqrt{ab}}\,\tan^{-1} \left(e^{mx}\sqrt{\frac{a}{b}}\,\right)$$
|
I don't know of the article where this result comes from, but it is definitely doable without it! First, notice that equality will fail if $a$ or $b$ are equal to zero. You cannot calculate the integral at all if both $a$ and $b$ equal zero. But the integral is very easy to solve if either $a$ or $b$ is zero. If both are non-zero, then you can pattern match what you have to something that we can integrate. Remember that $$\int \frac{dy}{1+y^2} = \arctan(y)+C$$ for any function $y$. Let's try to rewrite your integral to look like the above integral. $$\int{\frac{1}{ae^{mx}+be^{-mx}}}dx = \int \frac{1}{be^{-mx}}\left(\frac{1}{\frac{a}{b}e^{2mx}+1}\right)dx \\ =\frac{1}{b}\int \frac{e^{mx}}{\left(\sqrt{\frac{a}{b}e^{2mx}}\right)^2+1}dx \\ =\frac{1}{b}\int \frac{e^{mx}}{\left(\sqrt{\frac{a}{b}}e^{mx}\right)^2+1}dx$$ Introduce $$y = \sqrt{\frac{a}{b}}e^{mx}$$ to obtain $$dy = m\sqrt{\frac{a}{b}}e^{mx}dx \implies dx = \frac{1}{m}\sqrt{\frac{b}{a}}e^{-mx}dy$$ Hence $$\frac{1}{b}\int \frac{e^{mx}}{\left(\sqrt{\frac{a}{b}}e^{mx}\right)^2+1}dx = \frac{1}{b}\int \frac{e^{mx}}{\left(y\right)^2+1} \left(\frac{1}{m}\sqrt{\frac{b}{a}}e^{-mx}dy \right) \\ = \frac{1}{m}\sqrt{\frac{b}{a}}\cdot \frac{1}{b}\int \frac{e^{mx}\cdot e^{-mx}}{\left(y\right)^2+1}dy \\ = \frac{1}{m\sqrt{ab}}\int \frac{1}{\left(y\right)^2+1}dy$$ The integral has successfully been rewritten into a form that we recognize and can integrate. Thus, $$ \frac{1}{m\sqrt{ab}}\int \frac{1}{\left(y\right)^2+1}dy= \frac{1}{m\sqrt{ab}}\left(\arctan(y)+C \right)\\ = \frac{1}{m\sqrt{ab}}\arctan(y)+\tilde{C} \\ = \frac{1}{m\sqrt{ab}}\arctan\left(\sqrt{\frac{a}{b}}e^{mx} \right)+\tilde{C} $$
|
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|
How to find the value of this summation equation? The question is:
$$\sum_{i=1}^n (i^2+3i+4)$$
I get that
$$\sum_{i=1}^n i^2 = \frac{n(n+1)(n+2)}{6}$$ and $$3\sum_{i=1}^n i = \frac{3n(n+1)}{2}$$ so one would get
I'll call this form1: $$\frac{n(n+1)(n+2)}{6} + \frac{3n(n+1)}{2} + 4n$$
However, the textbook that I using says the answer is:
I'll call this form2: $$\frac{n(n^2+6n+17)}{3}$$
So the part I am confused with is the steps in between form1 and form2.
On a last note it been a good year since I've done any calculus so it would appreciated if you would point the relevant concepts so I can review. Thanks.
|
$$\sum_{k\le n} c=cn\neq c$$
In your example, $\displaystyle\sum 4=4n,$ yet you wrote $4$.
EDIT:
$$\frac{n(n+1)(n+2)}{6} + \frac{3n(n+1)}{2} + 4n = n(n^2+6n+17)/3$$
$$\frac{n(n+1)(n+2)}{2} + \frac{9n(n+1)}{2} + 12n = n^3+6n^2+17n$$
$$\frac{n^3+3n^2+2n}{2} + \frac{9n^2+9n}{2} = n^3+6n^2+5n$$
$${n^3+3n^2+2n} + {9n^2+9n}= 2n^3+12n^2+10n$$
$${2n} + {9n}= n^3+10n$$
$$1\neq n^2$$
Therefore one of them is incorrect.
We see that you messed up on $\sum i^2$, it equals $\frac{n(n+1)(2n+1)}{6}$
|
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|
How to prove that $\left(\sqrt{3}\sec{\frac{\pi}{5}}+\tan{\frac{\pi}{30}}\right)\tan{\frac{2\pi}{15}}=1$ From this geometry problem, I can not find geometry solution.
However the answer is $X=\frac{2\pi}{15}$ by geometry method.
Then I get the identity $$\left(\sqrt{3}\sec{\frac{\pi}{5}}+\tan{\frac{\pi}{30}}\right)\tan{\frac{2\pi}{15}}=1.$$
How to prove it by trigonometric method ?
Thank in advances.
|
This is not a fully sketched answer.I enjoyed while going through it and hence thought others might like it.
Since $\displaystyle\sin\frac{\pi}{10}=\frac{1}{4}(\sqrt{5}-1)$ we obtain $\displaystyle\sin\frac{\pi}{30}=\frac{1}{8}(\sqrt{30-6\sqrt{5}}-1-\sqrt{5})$ and consequently $$\tan\frac{\pi}{30}=\sqrt{7-2\sqrt{5}-2\sqrt{15-6\sqrt{5}}}$$
Now since $\displaystyle\cos\frac{\pi}{5}=\frac{1}{4}(\sqrt{5}+1)$ we obtain $\displaystyle\cos\frac{\pi}{15}=\frac{1}{8}(\sqrt{30+6\sqrt{5}}-1+\sqrt{5})$ and consequently $$\tan\frac{\pi}{15}=\frac{1}{\sqrt{7+2\sqrt{5}+2\sqrt{15+6\sqrt{5}}}}$$ which in turn implies $$\tan\frac{2\pi}{15}=\frac{\sqrt{7+2\sqrt{5}+2\sqrt{15+6\sqrt{5}}}}{3+\sqrt{5}+\sqrt{15+6\sqrt{5}}}$$
Finally note that $\displaystyle\cos\frac{\pi}{5}=\frac{1}{4}(\sqrt{5}+1)$. Putting all these together and after some good bit of time spent on simplification we obtain the result.
|
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|
Show that x is independent of this trig expression For some reason I cannot get the solution.
Show that $x$ is independent of: $\sin^2(x+y)+\sin^2(x+z)-2\cos(y-z)\sin(x+y)\sin(x+z)$
I have used all the identities but seem to be missing something.
|
$\sin^2(x+y)+\sin^2(x+z) =1-[\cos^2(x+y)-\sin^2(x+z)]$
Using Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$,
$\cos^2(x+y)-\sin^2(x+z)=\cos(2x+y+z)\cos(y-z)$
and then use Werner formula, for $2\sin(x+y)\sin(x+z)$
|
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|
How can we say two algebraic expressions are "equal" if one is undefined at certain points and the other isn't? I'm trying to understand why it is that we can say $\frac{x^2-1}{x-1} = \frac{(x-1)(x+1)}{(x-1)} = x+1$ but then have it also be the case that the two functions $f(x) = \frac{x^2-1}{x-1}$ and $g(x)=x+1$ are not equal, since $f$ is undefined when $x=1$.
That is, if it is true that $\frac{x^2-1}{x-1} = x+1$ then why can we not just simplify and say that $f(x)=x+1$ so that $f(1) = (1) + 1 = 2$ ?
|
we have $$\frac{x^2-1}{x-1}=\frac{(x-1)(x+1)}{x-1}=x+1$$ if $x \ne 1$
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "6",
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|
Convert the Polar Equation to Cartesian Coordinates $$
r^2=\sec 4\theta
$$
I graphed this equations using Wolfram Alpha and found it to be 2 hyperbolas. I'm having difficulty showing this using the standard equations
$$
x=r\cos\theta \;, \; y=r\sin\theta \;, and \; x^2 +y^2 =r^2
$$
My work so far:
$$
r^2 = \sec4\theta=\frac{1}{\cos4\theta}=\frac{1}{\cos(2\theta+2\theta)}=\frac{1}{\cos2\theta \cos2\theta - \sin2\theta\sin2\theta} \\ \\ =\frac{1}{1-8\sin^2\theta + 8\sin^4\theta}
$$
I'm getting nowhere from here. I've tried using a few other trig identities, but no luck! Can one please point me in the right direction? I would appreciate any help. Thank you!!!
|
i can take $$r^2 =\frac{1}{1-8\sin^2\theta + 8\sin^4\theta}$$ and turn it into a cartesian equation.
$\begin{align}
1 &=\frac{r^2}{r^4-8r^4\sin^2\theta + 8r^4\sin^4\theta}\\
&=\dfrac{(x^2+y^2)}{(x^2+y^2)^2 - 8(x^2+y^2)y^2 +8y^4}\\
&= \dfrac{(x^2+y^2)}{x^4 + y^4-6x^2y^2}\\
\end{align}$
|
{
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
find limit by rationalizing? I'm extremely new to calculus so please excuse my lack of lingo/formatting.!
Here is the problem:
$$\lim_{n\to\infty} \sqrt{5n^2 + 4n +2} - \sqrt{5n^2 - 2n - 1}$$
|
To rationalize:
*
*Multiply by $$\frac{\sqrt{5n^2 + 4n +2} + \sqrt{5n^2 - 2n -
1}}{\sqrt{5n^2 + 4n +2} + \sqrt{5n^2 - 2n - 1}}$$
$$\left(\sqrt{5n^2 + 4n +2} - \sqrt{5n^2 - 2n -
1}\right)\cdot\frac{\sqrt{5n^2 + 4n +2} + \sqrt{5n^2 - 2n -
1}}{\sqrt{5n^2 + 4n +2} + \sqrt{5n^2 - 2n - 1}}$$ $$= \frac{6n+3}{\sqrt{5n^2 + 4n +2} + \sqrt{5n^2 - 2n - 1}}$$
*Now, divide numerator and denominator by $n$.
You should end up with a limit $$\frac{6}{2\sqrt 5} = \frac 3{\sqrt 5} = \frac {3\sqrt 5}{5}$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Solve for $y$ in $x=\sqrt{(y-1)/(y+1)}$ I always struggle with this:
Express $y$ in terms of $x$ where
$$x = \sqrt\frac{y-1}{y+1}$$
I know to square both sides and get $x^2 = \frac{y-1}{y+1}$
Then I'm thinking multiply both sides by $y+1?$ So we get $(y+1)x^2 = y-1$
But where do I go from here? If there was only one y I would be fine.
|
Here is what I would do:
\begin{align}
x=\sqrt{\frac{y-1}{y+1}} &\Longleftrightarrow x^2 = \frac{y-1}{y+1}\\[1em]
&\Longleftrightarrow (y+1)x^2 = y-1\\[1em]
&\Longleftrightarrow yx^2+x^2=y-1\\[1em]
&\Longleftrightarrow yx^2-y=-(x^2+1)\\[1em]
&\Longleftrightarrow y(x^2-1)=-(x^2+1)\\[1em]
&\Longleftrightarrow y = \frac{-1}{-1}\cdot \frac{x^2+1}{1-x^2}\\[1em]
&\Longleftrightarrow y = \frac{x^2+1}{1-x^2}
\end{align}
Make sure you do not forget about the value restricts though (i.e., $x$ cannot be negative, $y\neq -1$, etc.).
|
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|
Closed form of a recursive relation A sequence $\langle a_n\rangle$ is defined recursively by $a_1=0$, $a_2=1$ and for $n\ge 3$,
$$a_n=\frac 12 na_{n-1}+\frac 12n(n-1)a_{n-2}+(-1)^n\left(1-\frac n2\right).$$
Find a closed form expression for
$$f_n=a_n+2\binom n1a_{n-1}+3\binom n2a_{n-2}+\cdots+(n-1)\binom n{n-2}a_2+n\binom n{n-1}a_1.$$
I substituted $b_k=\binom n{n-k}a_k$ which reduces the given recursion to $$b_n=\frac {b_{n-1}}2+b_{n-2}+(-1)^n\left(1-\frac n2\right)\\ \Longrightarrow2b_n-b_{n-1}-2b_{n-2}=(-1)^n(2-n)\\ \Longrightarrow 2b_{n-1}-b_{n-2}-2b_{n-3}=-(-1)^n(3-n)\\ \Longrightarrow2b_{n-1}-b_{n-2}-2b_{n-3}=-(-1)^n(3-n)\\ \Longrightarrow 2b_{n-2}-b_{n-3}-2b_{n-4}=(-1)^n(4-n)$$ Adding the last four equations, we get $$2b_n+3b_{n-1}-2b_{n-2}-5b_{n-3}-2b_{n-4}=0$$ Now using the standard way of solving such recursions, we set $b_k=\lambda^k$, which gives $$2\lambda^4+3\lambda^3-2\lambda^2-5\lambda-2=0$$ We have to find $f_n=a_n+2\binom n1a_{n-1}+3\binom n2a_{n-2}+\cdots+(n-1)\binom n{n-2}a_2+n\binom n{n-1}a_1\\=b_n+2b_{n-1}+3b_{n-2}+\cdots+(n-1)b_2+nb_1\\=\lambda^n+2\lambda^{n-1}+3\lambda^{n-2}+\cdots+(n-1)\lambda^2+n\lambda$
This is where I got stuck. What should I do after this?
|
With Abhishek Bakshi hint ,Now I have complete this answer,
since
$$\dfrac{a_{n}}{n!}=\dfrac{1}{2}\dfrac{a_{n-1}}{(n-1)!}+\dfrac{1}{2}\dfrac{a_{n-2}}{(n-2)!}+\dfrac{(-)^n(1-\dfrac{n}{2})}{n!}$$
let
$b_{n}=\dfrac{a_{n}}{n!},b_{1}=0,,b_{2}=\dfrac{1}{2} $then we have
$$b_{n}=\dfrac{1}{2}b_{n-1}+\dfrac{1}{2}b_{n-2}+\dfrac{(-)^n(1-\dfrac{n}{2})}{n!}$$
$$\Longrightarrow b_{n}-b_{n-1}=-\dfrac{1}{2}(b_{n-1}-b_{n-2})+\dfrac{(-)^n(1-\dfrac{n}{2})}{n!}$$
let
$b_{n}-b_{n-1}=c_{n},c_{2}=\dfrac{1}{2}$,then we have
$$2c_{n}+c_{n-1}=\dfrac{2(-1)^n}{n!}+\dfrac{(-1)^n}{(n-1)!}$$
so
$$c_{n}=\dfrac{(-1)^n}{n!}+C$$
since
$c_{2}=\dfrac{1}{2}\Longrightarrow C=0,$,so
$$c_{n}=\dfrac{(-1)^n}{n!}\Longrightarrow b_{n}=b_{n-1}+\dfrac{(-1)^n}{n!}$$
and since
$$f_{n}=a_{n}+2\binom{n}{1}a_{n-1}+\cdots+(n-1)\binom{n}{n-2}a_{2}+n\binom{n}{n-1}a_{1},a_{1}=0$$
so
\begin{align*}\dfrac{f_{n}}{n!}&=\dfrac{a_{n}}{n!}+2\dfrac{b_{n-1}}{(n-1)!}+\cdots+\dfrac{(n-1)}{(n-2)!}\dfrac{a_{2}}{2!}\\
&=b_{n}+2b_{n-1}+\dfrac{3}{2!}b_{n-2}+\cdots+\dfrac{n-1}{(n-2)!}b_{2}
\end{align*}
so we have
\begin{align*}\dfrac{f_{n+1}}{(n+1)!}-\dfrac{f_{n}}{n!}&=\left(b_{n+1}+2b_{n}+\cdots+\dfrac{n}{(n-1)!}b_{2}\right)-\left(b_{n}+2b_{n-1}+\cdots+\dfrac{n-1}{(n-2)!}b_{2}\right)\\
&=(b_{n+1}-b_{n})+2(b_{n}-b_{n-1})+\dfrac{3}{2!}(b_{n-1}-b_{n-2})+\cdots\\
&=\dfrac{(-1)^{n+1}}{n+1}+\dfrac{2(-1)^n}{n!}+\dfrac{3(-1)^{n-1}}{2!(n-1)!}+\cdots+\dfrac{n}{2!(n-1)!}
\end{align*}
let $$A_{n}=\dfrac{(-1)^{n+1}}{n+1}+\dfrac{2(-1)^n}{n!}+\dfrac{3(-1)^{n-1}}{2!(n-1)!}+\cdots+\dfrac{n}{2!(n-1)!}$$
so we have
$$(-1)^{n+1}(n+1)!A_{n}=1-2\binom{n+1}{1}+3\binom{n+1}{2}+\cdots+(-1)^{n-1}n\binom{n+1}{n-1}$$
Note
\begin{align*}(1-1)^{n+1}&=1-\binom{n+1}{1}+\binom{n+1}{2}+\cdots+(-1)^{n-1}\binom{n+1}{n-1}+(-1)^n(n+1)+(-1)^{n+1}\\
&=0
\end{align*}
and use well know indentity
$$k\binom{n+1}{k}=(n+1)\binom{n}{k-1}$$
so
\begin{align*}
&-\binom{n+1}{1}+2\binom{n+1}{2}+\cdots+(-1)^{n+1}(n+1)\binom{n+1}{n+1}\\
&=-(n+1)[\binom{n}{0}-\binom{n}{1}+\cdots+(-1)^n\binom{n}{n}]\\
&=0
\end{align*}
so we have
$$(-1)^{n+1}(n+1)!A_{n}=(-1)^{n+1}(n^2+n-1)$$
so
$$\dfrac{f_{n+1}}{(n+1)!}-\dfrac{f_{n}}{n!}=\dfrac{n^2+n-1}{(n+1)!}=\dfrac{n+1}{n!}-\dfrac{n+2}{(n+1)!}$$
so
$$\dfrac{f_{n}}{n!}=C-\dfrac{n+1}{n!}$$,note $f_{1}=0\Longrightarrow C=2$
so
$$f_{n}=2n!-(n+1)$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
the general solution of $y(n+3)-\frac{2}{3}y(n+1)+\frac{1}{3}y(n) = 0$ I have some trouble finding the correct solution for the difference equation
$$y(n+3)-\frac{2}{3}y(n+1)+\frac{1}{3}y(n) = 0$$
I've found that the characteristic equation of the difference equation is $\lambda^3-\frac{2}{3}\lambda+\frac{1}{3}$.
By computation I then have;
\begin{array}{lcl}
\lambda^3-\dfrac{2}{3}\lambda+\dfrac{1}{3}\\
(\lambda+1)(\lambda^2-\lambda+\dfrac{1}{3})\\
\lambda_1 = -1, \lambda_2 = \dfrac{1-i\sqrt{\frac{1}{3}}}{2}, \lambda_3 = \dfrac{1+i\sqrt{\frac{1}{3}}}{2}
\end{array}
I checked my answer using wolframalpha and it gave me the eigenvalues
$$\begin{array}{lcl}
\lambda_1 = -1, \lambda_2 = \dfrac{1-i\sqrt{3}}{2}, \lambda_3 = \dfrac{1+i\sqrt{3}}{2}
\end{array}$$
Could someone please tell me what I did wrong, or what I should do differently?
|
How did you input the equation into Wolfram Alpha?
You have as solutions
$$
\begin{align}
a_n&=(-1)^n\tag{1}\\[9pt]
b_n
&=\frac12\left[\,\left(\frac{1+\frac i{\sqrt{3}}}{2}\right)^n+\left(\frac{1-\frac i{\sqrt{3}}}{2}\right)^n\,\right]\\
&=\mathrm{Re}\left[\,\left(\frac{1+\frac i{\sqrt{3}}}{2}\right)^n\,\right]\tag{2}\\[9pt]
c_n
&=\frac1{2i}\left[\,\left(\frac{1+\frac i{\sqrt{3}}}{2}\right)^n-\left(\frac{1-\frac i{\sqrt{3}}}{2}\right)^n\,\right]\\
&=\mathrm{Im}\left[\,\left(\frac{1+\frac i{\sqrt{3}}}{2}\right)^n\,\right]\tag{3}
\end{align}
$$
and any linear combination of $a_n$, $b_n$, and $c_n$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
$\sum_1^n 2\sqrt{n} - \sqrt{n-1} - \sqrt{n+1} $ converge or not? how to check if this converge? $$\sum_{n=1}^\infty a_n$$
$$a_n = 2\sqrt{n} - \sqrt{n-1} - \sqrt{n+1}$$
what i did is to show that:
$$a_n =2\sqrt{n} - \sqrt{n-1} - \sqrt{n+1} > 2\sqrt{n} - \sqrt{n+1} - \sqrt{n+1} = 2\sqrt{n} - 2\sqrt{n+1} = -2({\sqrt{n+1} - \sqrt{n}}) = -2(({\sqrt{n+1} - \sqrt{n}})\frac{\sqrt{n+1} + \sqrt{n}}{\sqrt{n+1} + \sqrt{n}}) = -2\frac{1}{\sqrt{n+1} + \sqrt{n}} > -2\frac{1}{2\sqrt{n+1}} = \frac{-1}{\sqrt{n+1}} = b_n $$
and we know that:
$$\sum_{n=1}^\infty b_n$$
doesnt converge cause
$$\sum_{n=1}^\infty \frac{1}{\sqrt{n}}$$
doesnt converge
so from here my conclusion is that $\sum_{n=1}^\infty a_n$ doesnt converge
but i know the final answer is that it does converge
so what am i doing wrong?
|
We have
$$2\sqrt{n} - \sqrt{n-1} - \sqrt{n+1}=\sqrt n\left(2-\sqrt{1-\frac1n}-\sqrt{1+\frac1n}\right)\\\sim_\infty \sqrt n\left(2-1+\frac1{2n}+\frac1{4n^2}-1-\frac1{2n}+\frac1{4n^2}\right)=\frac1{2n^{3/2}}$$
so the given series is convergent by comparison with a convergent Riemann series.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Overdetermined system with parameters $$\begin{cases}
ax & +y &=2a\\
x &+by &= b \\
ax &+ (5-b^2)y &= 1 \tag{A,B,C (in order)}
\end{cases}$$
Where $(x,y)$ are variables and $a,b$ are constants.
*
*What combination of $a$ and $b$ yields infinitely many solutions to our system?
By subtracting $A$ from $C$ we get:
$\begin{cases}
ax & +y &=2a\\
x &+by &= b \\
&(5-b^2)y - y &= 1 - 2a \tag{A',B',C' (in order)}
\end{cases}$
$\iff\begin{cases}
ax & +y &=2a\\
x &+by &= b \\
&y(4-b^2) &= 1 - 2a \tag{A'',B'',C'' (in order)}
\end{cases}$
$\iff\begin{cases}
ax & +y &=2a\\
x &+by &= b \\
&y &= \frac{1 - 2a}{4-b^2} \tag{A''',B''',C''' (in order)}
\end{cases}$
By subtracting $C'''$ from $A'''$ we get:
$$\begin{cases}
ax &&=2a - \frac{1 - 2a}{4-b^2}\\
x &+by &= b \\
&y &= \frac{1 - 2a}{4-b^2}
\end{cases}$$
$$\iff\begin{cases}
x &&=2 - \frac{1 - 2a}{4a-ab^2}\\
x &+by &= b \\
&y &= \frac{1 - 2a}{(4-b^2)}
\end{cases}$$
Substituting $x$ and $y$ into $B$ gets us the following expression:
$$\left(2-\frac{1-2a}{4a-ab^2}\right) + b\cdot\left(\frac{1-2a}{4-b^2}\right) = b$$
Did I do anything right? Is this leading to the solution? Is there any more examples of this particular problem anywhere? Can't seem to find any.
NB! I ONLY SEEK THE VALUES OF $a$ AND $b$ WHEN THEY YIELD INFINITELY MANY SOLUTIONS TO OUR SYSTEM
|
My advice is to avoid dividing by variable expressions (since we might lose solutions or otherwise be implicitly making certain assumptions). We can row reduce the corresponding augmented matrix to get:
$$
\left[\begin{array}{cc|c}
a & 1 & 2a \\
1 & b & b \\
a & 5 - b^2 & 1
\end{array}\right]
\sim
\left[\begin{array}{cc|c}
1 & b & b \\
0 & 1-ab & 2a-ab \\
0 & 4 - b^2 & 1 - 2a
\end{array}\right]
$$
Now since the system must have infinitely many solutions, the second column cannot have a pivot. Thus, $1 - ab = 4 - b^2 = 0$. But then since the system must be consistent, the last two entries in the augmented column must be zero, so $2a - ab = 1 - 2a = 0$. Since:
$$
1 - 2a = 0 \iff \boxed{a = 1/2}
$$
we can substitute this into $1 - ab = 0$ to get that $\boxed{b = 2}$.
Indeed, this combination of $a$ and $b$ values makes the last two rows of the augmented matrix all zeroes, as desired. $~~\blacksquare$
|
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|
Why it's true? $\arcsin(x) +\arccos(x) = \frac{\pi}{2}$ The following identity is true for any given $x \in [-1,1]$:
$$\arcsin(x) + \arccos(x) = \frac{\pi}{2}$$
But I don't know how to explain it.
I understand that the derivative of the equation is a truth clause, but why would the following be true, intuitively?
$$\int^{x}_{C1}\frac{1\cdot dx}{\sqrt{1-x^{2}}} + \int^{x}_{C2}\frac{-1 \cdot dx}{\sqrt{1-x^{2}}} =\\
\arcsin(x) - \arcsin(C1) + \arccos(x) - \arccos(C2) = 0 \\
\text{while } \arcsin(C1) + \arccos(C2) = \frac{\pi}{2}$$
I can't find the right words to explain why this is true?
Edit #1 (25 Jan, 20:10 UTC):
The following is a truth clause:
$$
\begin{array}{ll}
\frac{d}{dx}(\arcsin(x) + \arccos(x)) = \frac{d}{dx}\frac{\pi}{2} \\
\\
\frac{1}{\sqrt{1-x^{2}}} + \frac{-1}{\sqrt{1-x^{2}}} = 0
\end{array}
$$
By integrating the last equation, using the limits $k$ (a constant) and $x$ (variable), I get the following:
$$
\begin{array}{ll}
\int^x_k\frac{1}{\sqrt{1-x^{2}}}dx + \int^x_k\frac{-1}{\sqrt{1-x^{2}}}dx = \int^x_k0 \\
\\
\arcsin(x) - \arcsin(k) + \arccos(x) - \arccos(k) = m \text{ (m is a constant)}\\
\\
\arcsin(x) + \arccos(x) = m + \arcsin(k) + \arccos(k) \\
\\
\text{Assuming that } A = m + \arcsin(k) + \arccos(k) = \frac{\pi}{2} \text{ ,for } x \in [-1,1]
\end{array}
$$
Using Calculus, why is that true for every $x \in [-1,1]$?
Edit #2:
A big mistake of mine was to think that $\int^x_k0 = m \text{ (m is const.)}$, but that isn't true for definite integrals.
Thus the equations from "Edit #1" should be as follows:
$$
\begin{array}{ll}
\int^x_k\frac{1}{\sqrt{1-x^{2}}}dx + \int^x_k\frac{-1}{\sqrt{1-x^{2}}}dx = \int^x_k0 \\
\\
\arcsin(x) - \arcsin(k) + \arccos(x) - \arccos(k) = 0\\
\\
\arcsin(x) + \arccos(x) = \arcsin(k) + \arccos(k) \\
\\
A = \arcsin(k) + \arccos(k) = \frac{\pi}{2} \text{ ,for } x \in [-1,1]
\end{array}
$$
|
By definition, $\arcsin(x)$ is the angle $\alpha$ such that $\sin(\alpha) = x$ and $-\pi/2 \le \alpha \le \pi/2$, while $\arccos(x)$ is the angle $\beta$ such that $\cos(\beta) = x$ and $0 \le \beta \le \pi$. Since $-\pi/2 \le \alpha \le \pi/2$, $\cos(\alpha) \ge 0$, so we have $\cos(\alpha) = \sqrt{1 - x^2}$. Similarly $\sin(\beta) = \sqrt{1-x^2}$. Now
$$\eqalign{-\pi/2 &\le \arcsin(x) + \arccos(x) = \alpha + \beta \le 3 \pi/2 \cr\sin(\alpha + \beta) &= \sin(\alpha) \cos(\beta) + \cos(\alpha) \sin(\beta) =
x^2 + 1 - x^2 = 1\cr
\cos(\alpha + \beta) &= \cos(\alpha) \cos(\beta) - \sin(\alpha)\sin(\beta) = \sqrt{1-x^2} x - x \sqrt{1-x^2} = 0}$$
and the only angle in this interval with that sine and cosine is $\pi/2$.
|
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|
How to divide trigonometric ratios using identities? $$\frac{1-\tan^2x}{1+\tan^2x}$$
We know:
$$\frac{1-\frac{\sin^2x}{\cos^2x}}{1+\frac{\sin^2x}{\cos^2x}}$$
Now what? Flip denominator and times numerator?
Which equals ???
Please help - Thanks
|
$$
\tan^2 x + 1 = \sec^2 x
$$
Thus
$$
\frac{1-\tan^2 x}{\sec^2 x} = \cos^2x - \sin^2x
$$
But going from your point of view
$$
\frac{1-\frac{\sin^2x}{\cos^2x}}{1+\frac{\sin^2x}{\cos^2x}}= \frac{\frac{1}{\cos^2x}}{\frac{1}{\cos^2x}}\frac{\cos^2-\sin^2x}{\cos^2x+\sin^2x}=\\
\frac{\cos^2-\sin^2x}{\cos^2x+\sin^2x}
$$
then use the fact
$$
\cos^2x+\sin^2x=1
$$
Dividing through by $\cos^2x$ and knowing that $\frac{1}{\cos x} =\sec x$ will lead to the first identity.
|
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|
Prove that distinct Fermat Numbers are relatively prime The Fermat numbers are defined by $F_m = 2^{2^m} + 1$.
Prove that for $m \ne n$ we have $(F_m, F_n) = 1$.
I have to first prove that $F_{m+1} = F_0F_1 \cdots F_m + 2$ by representing $F_{m+1}$ in terms of $F_m$.
|
To see that $F_{m+1} = F_0 F_1\cdots F_m + 2$, proceed as follows.
\begin{align}F_{m+1} &= (2^{2^m} - 1) + 2\\
&= [(2^{2^{m-1}})^2 - 1] + 2\\
&= (2^{2^{m-1}} - 1)F_m + 2\\
&= [(2^{2^{m-2}})^2 - 1]F_m + 2\\
&= (2^{2^{m-2}} - 1)F_{m-1}F_m + 2\\
&\ldots\\
&= (2^{2^0} - 1)F_1F_2\cdots F_m + 2\\
&= F_0F_1\cdots F_m + 2.
\end{align}
Now to prove the theorem, let $d = \gcd(F_m,F_n)$, and assume, without loss of generality, that $m > n$. Then $d|F_m$ and $d|F_n$, which implies $d|F_0\cdots F_{m-1}$, since $F_n$ is a factor of $F_0\cdots F_{m-1}$ when $n < m$. Therefore, $d|(F_m - F_0\cdots F_{m-1})$, or $d|2$. Since $d$ is an odd positive number, the condition $d|2$ implies $d = 1$.
|
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How prove $\sum_{i=2}^na_i^{1-\frac{1}{i}} < S+2\sqrt{S}$ for $S=a_2+\dots +a_n.$ Let $a_2,\dots,a_n>0$ and $S=a_2+\dots +a_n.$ How prove $\sum_{i=2}^na_i^{1-\frac{1}{i}} < S+2\sqrt{S}.$
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Because $\sum_{i=1}^na_i^{1-\frac{1}{i}}=\sum_{i=2}^na_i^{1-\frac{1}{i}}+1 < S+2\sqrt{S}+1=(\sqrt S +1)^2 \Leftrightarrow \sum_{i=2}^na_i^{1-\frac{1}{i}} < S+2\sqrt{S}.$
$ \ $
It's sufficient to show $(\sqrt S +1)^2 > {\sum_{i=1}^na_i^{1-\frac{1}{i}}}\tag 1 $
Suppose $S_n=a_2+\dots +a_n.$
Prove by induction:
When $n=1,2$ (1) holds obviously.
If $m=n-1\ge 2$ holds,then for $m=n$:
$(\sqrt S_{n-1} + 1)^2 +a_n^{1-\frac 1n}=S_{n-1}+1+2\sqrt S_{n-1}+a_n^{1-\frac 1n}
\\ =S_n+1+2\sqrt S_{n-1}<S_n+1+2\sqrt S_{n}=(\sqrt S_n +1)^2$
Hence $\sum_{i=1}^{n}a_i^{1-\frac{1}{i}} < (\sqrt S_{n-1} + 1)^2+a_n^{1-\frac 1n} < (\sqrt S_n +1)^2$
|
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Combinatorics, surjective functions with conditions Question: $A=\left\{ 1,2,3,4,5\right\} $
, $B=\left\{ 1,2,3\right\} $
. How many surjective functions are there such that $ f(1)\neq1$
,$f(2)\neq2$
,$ f(3)\neq2$
.
Solution: Overall we have $3^{5}-{3 \choose 1}2^{5}+{3 \choose 2}1^{5}=150$
functions that is surjective. We will denote $A_{i,j}=\left\{ f:A\rightarrow B\,|\, f\,\text{is surjective and }f(i)=j\right\} $
We will calculate $\left|A_{1,1}\cup A_{2,2}\cup A_{3,2}\right|$
then subtract it from 150.
So $\left|A_{1,1}\right|=\left|A_{2,2}\right|=\left|A_{3,2}\right|=3^{4}-{3 \choose 1}2^{4}+{3 \choose 2}1^{4}=36\Rightarrow\left|A_{1,1}\right|+\left|A_{2,2}\right|+\left|A_{3,2}\right|=3.36=108$
and $\left|A_{1,1}\cap A_{2,2}\right|=\left|A_{1,1}\cap A_{3,2}\right|=\left|A_{2,2}\cap A_{3,2}\right|=3^{3}-{3 \choose 1}2^{3}+{3 \choose 2}1^{3}=6\Rightarrow\left|A_{1,1}\cap A_{2,2}\right|+\left|A_{1,1}\cap A_{3,2}\right|+\left|A_{2,2}\cap A_{3,2}\right|=3.6=18$
lastly $\left|A_{1,1}\cap A_{2,2}\cap A_{3,2}\right|=3.2=6$
because we have 4 or 5 has to go to 3, otherwise we wouldn't have a surjective functions.
So my answer was $150-108+18-6=54$
. However I was told that the solution was 45. I would appreciate if someone can tell me where my mistake is. Thank you!
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As an alternate method, let $A_i$ be the preimage of $i$ for $1\le i\le3$.
$\textbf{1)}$ If $1\in A_2$,
then there are $3\cdot3=9$ possible distributions of 4 and 5 if $2\in A_i$ and $3\in A_j$ with $i\ne j$
and there are $3^2-2^2=5$ possible distributions of 4 and 5 if $2, 3 \in A_i$ for $i=1$ or $i=3$.
$\textbf{2)}$ If $1\in A_3$,
then there are $3^2-2^2=5$ possible distributions of 4 and 5 if $2\in A_1$ or $3\in A_1$
and there are $2\cdot1=2$ possible distributions of 4 and 5 if $2,3\in A_3$.
Thus there are $2\cdot9+2\cdot5+3\cdot5+1\cdot2=45$ such functions.
|
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Prove that there are no positive integers $a, b$ and $n >1$ such that $a^n – b^n$ divides $ a^n + b^n$. Prove that there are no positive integers $a$ , $b$ and $n>1$ such that $a^{n}–b^{n}$ divides $a^{n}+b^{n}$.
Can someone provide me a proof of this and explain it to me please.
|
This is a problem that appears in this number theory book by Nivens et al (1.2.46, p 19). The following is a solution readily attained.
Problem: Prove that there are no positive integers $a,b,n > 1$ such that $(a^n-b^n)\mid (a^n+b^n).$
Proof. We may first prove the following fact:
$$
\text{For the integers $x,y$ such that $(x,y)=1, (x+y,x-y)=1$ or $2$}.
$$
Since $(x,y)=1$, there is an integer $s,t$ such that $sx+ty=1$. Then
$$
(s+t)(x+y)+(s-t)(x-y)=2sx+2ty=2.
$$
Hence, $(x+y,x-y)\leq 2$ (by elementary number theory facts concerning $\mathrm{gcd}$); thus, we just proved the fact above.
Now, suppose that there are positive integers $a,b$ and $n>1$ such that $(a^n-b^n)\mid (a^n+b^n)$. Note that we may assume $a>b$ without loss of generality. Then there is a positive integer $Q$ which satisfies
$$
(a^n+b^n)\mid Q(a^n-b^n).
$$
Let $(a,b)=d$. Then divide this equation by $d^n$ to get that
$$
\left(\left(\frac{a}{d}\right)^n-\left(\frac{b}{d}\right)^n\right) = Q\left(\left(\frac{a}{d}\right)^n+\left(\frac{b}{d}\right)^n\right)
$$
and $\left(\left(\frac{a}{d}\right),\left(\frac{b}{d}\right)\right)=1$. Therefore, we can assume that there are positive integers $a,b$ which are relatively prime with $a>b$ and $n>1$ such that $(a^n-b^n)\mid (a^n+b^n)$. Now, $(a,b)=1$ implies that $(a^n,b^n)=1$, so we get that
$$
a^n-b^n = (a^n-b^n,a^n+b^n)=1\;\text{or}\;2,
$$
where first equality holds by $(a^n-b^n)\mid (a^n+b^n)$ and second equality holds by the fact we first proved.
But $a^n-b^n=(a-b)(a^{n-1}+\cdot+b^{n-1})$ and it is trivial that $a\neq b$ to make $(a^n-b^n)\mid (a^n+b^n)$ sense. So, $a>b\geq 1$. Therefore, $(a-b)\geq 1$ and $a^{n-1}+\cdots+b^{n-1}\geq 2+1=3$, since $n > 1$. Then we get that
$$
a^n-b^n=(a-b)(a^{n-1}+\cdots+b^{n-1}\geq 3,
$$
a contradiction, because it cannot be $1$ or $2$. Therefore, there are no positive integers $a,b,n>1$ such that $(a^n-b^n)\mid (a^n+b^n)$. $\Box$
|
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|
existence of solution to congruence $x^4 \equiv -4 \pmod p$ I stuck with the following question:
For which $p$ (prime numbers) there is a solution for the following congruence:
$x^4 \equiv -4 \pmod p$
I would greatly appreciate any help
|
$$x^4+4=(x^2+2)^2-(2x)^2=(x^2+2x+2)(x^2-2x+2)$$
If $p$ divides both $(x^2+2x+2),(x^2-2x+2);$
$p$ must divide $(x^2+2x+2)-(x^2-2x+2)=4x$
If $p$ divides $4,p=2$
Else if $p|x\implies p$ must divide $-4\implies p=2$
So, for odd prime $p,$ it must divide exactly one of $x^2\pm2x+2=(x\pm1)^2+1$
So, we need $(x\pm1)^2\equiv-1\pmod p$
See $-1$ is a quadratic residue modulo $p$ if and only if $p\equiv 1\pmod{4}$
|
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|
Principal value of Fourier Integral I have tried to find the principal value of
$$\int_{-\infty}^\infty {\sin(2x)\over x^3}\,dx.$$
As $ {\sin(2x)\over x^3}$ is an even function, its integral may not be zero in the given limits. I cannot calculate its principal value as I met with a third order pole at $z=0$. The expected value is $-2\pi$. How can I compute the principal value?
|
Unfortunately $$PV\int_{-\infty}^\infty {\sin 2x\over x^3}\,dx$$ diverges.
The integrals $\int_\frac{\pi}{4}^{\infty} {\sin 2x\over x^3}\,dx=\int_{-\infty}^{-\frac{\pi}{4}}{\sin 2x\over x^3}\,dx$ exist.
Since $$\sin 2x\ge \frac{4}{\pi}x$$ for $0<x<\frac{\pi}{4}$, we have $$\frac{\sin 2x}{x^3}\ge \frac{4}{\pi}\frac{1}{x^2} $$ for $0<x<\frac{\pi}{4}$. Thus $$\int_{-\frac{\pi}{4}}^{-\epsilon} \frac{\sin 2x}{x^3} dx=\int_\epsilon^{\frac{\pi}{4}} \frac{\sin 2x}{x^3} dx\ge \int_\epsilon^{\frac{\pi}{4}} \frac{4}{\pi}\frac{1}{x^2} dx=\frac{4}{\pi}\left(\frac{1}{\epsilon}-\frac{4}{\pi}\right).$$
Therefore $$PV\int_{-\infty}^{\infty} \frac{\sin 2x}{x^3} dx =\lim_{ \epsilon\to 0} \left( \int_{-\infty}^{-\frac{\pi}{4}}
+\int_{-\frac{\pi}{4}}^{-\epsilon} +\int_\epsilon^{\frac{\pi}{4}}+\int_\frac{\pi}{4}^{\infty}\right) {\sin 2x\over x^3}\,dx=+\infty.$$
|
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|
Determine the set of points $z$ that satisfy the condition $|2z|>|1+z^2|$ Determine the set of points $z$ that satisfy the condition $|2z|>|1+z^2|$
I tried to redo this problem and got to this point
$|2z|>|1+z^2|$ $\Rightarrow$ $2|z|>1+|z^2|$ $\Rightarrow$ $2|z|>1+z\overline z$
Let $z=x+iy$ then
$$2\sqrt{x^2+y^2}>1+(x+iy)(x-iy)$$
$$2\sqrt{x^2+y^2}>1+(x^2+y^2)$$
$$0>1-2\sqrt{x^2+y^2}+(x^2+y^2)$$
$$0>(1-\sqrt{x^2+y^2})^2$$
since $x,y$ are real number , so there is no solution for this inequality?
|
let $D = \{z \colon 2|z| < |1 + z^2|\}.$ observe that the region $D$ is symmetric with respect to the transformations $z \to -z, z \to \bar z$ and $z \to \dfrac{1}{z}$ therefore it is enough to worry about the portion of $D$ in the first quadrant.
Let $o = 0, a = 1$ and $z = re^{it}, p = 1 + r^2e^{2it}$ so that $z$ is on the boundary of $D.$ applying rule of cosine to the triangle, we have $$\cos 2t = 2 - \frac{1}{2}\left( r^2 + \frac{1}{r^2} \right).$$
looking at the graph of $y = 2 - \frac{1}{2}(x + 1/x),$ you can verify that $-1 \le y \le 1$ for $3 - 2\sqrt 2 \le x \le 3 + 2\sqrt 2$ and $f(1) = 1, f(2 \pm \sqrt 3) = 0$
i plotted the region $D$ it contains the unit circle and has a hole(lens) containing the origin.
so part of $D$ in the first quadrant is given by
$$\{re^{it} \colon t = \frac{1}{2}\cos^{-1}\left( 2 - \frac{1}{2}( r^2 + \frac{1}{r^2} ) \right), \sqrt{3 - 2\sqrt 2} \le r \le \sqrt{3 + 2 \sqrt 2} \}.$$
|
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Find $\int_0^1 \frac{dx}{(1+x^n)^2\sqrt[n]{1+x^n}}$ Find $$\int_0^1 \frac{dx}{(1+x^n)^2\sqrt[n]{1+x^n}}$$
with $n \in \mathbb{N}$.
My tried:
I think that, I need to find the value of
$$I_1=\int_{0}^1 \frac{dx}{(1+x^n)\sqrt[n]{1+x^n}}$$
because:
$$\begin{align} I_1=\int_{0}^1 \frac{dx}{(1+x^n)\sqrt[n]{1+x^n}} &= \frac{x}{(1+x^n)\sqrt[n]{1+x^n}} \Bigg|_0^1-\int_0^1 x \ d(1+x^n)^{-1-\frac{1}{n}} \\ &=\frac{1}{2\sqrt[n]{2}}-\int_0^1 x\left(-1-\frac{1}{n}\right)(1+x^n)^{-2-\frac{1}{n}}(nx^{n-1}) \ dx \\ &=\frac{1}{2\sqrt[n]{2}}-\int_0^1(-n-1)\frac{x^n}{(1+x^n)^2\sqrt[n]{1+x^n}} \ dx\\ &=\frac{1}{2\sqrt[n]{2}}+(n+1)\int_0^1\frac{x^n+1-1}{(1+x^n)^2\sqrt[n]{1+x^n}} \ dx \\ &=\frac{1}{2\sqrt[n]{2}}+(n+1)I_1-(n+1)\int_0^1 \frac{dx}{(1+x^n)^2\sqrt[n]{1+x^n}}.\end{align} $$
So that,
$$ (n+1)\int_0^1 \frac{dx}{(1+x^n)^2\sqrt[n]{1+x^n}}=\frac{1}{2\sqrt[n]{2}}+nI_1$$
But how to find the following integral ?
$$I_1=\int_{0}^1 \frac{dx}{(1+x^n)\sqrt[n]{1+x^n}}$$
|
Calculation of $$I = \int\frac{1}{(1+x^n)\sqrt[n]{1+x^n}}dx\;,$$ Now Put $(1+x^n) = t^n\;,$ Then $nx^{n-1}=nt^{n-1}dt\Rightarrow x^{n-1}dx = t^{n-1}dt$
So we get $$I=\int\frac{(t^n-x^n)}{t^{n+1}}dx = \int \left[\frac{dx}{t}-\frac{dt}{t^2}\right] = \int d\left(\frac{x}{t}\right)=\frac{x}{t}+\mathcal{C}$$
So we get $$\int_{0}^{1}\frac{1}{(1+x^n)\sqrt[n]{1+x^n}}dx = \left[\frac{x}{\sqrt[n]{1+x^n}}\right]_{0}^{1} = \frac{1}{2^{\frac{1}{n}}}$$
|
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no. of real roots of the equation $ 1+\frac{x}{1}+\frac{x^2}{2}+\frac{x^3}{3}+............+\frac{x^7}{7} = 0$
The no. of real roots of the equation $\displaystyle 1+\frac{x}{1}+\frac{x^2}{2}+\frac{x^3}{3}+............+\frac{x^7}{7} = 0 $
$\bf{My\; Try::}$ First we will find nature of graph of function $\displaystyle 1+\frac{x}{1}+\frac{x^2}{2}+\frac{x^3}{3}+............+\frac{x^7}{7}$
So $$\displaystyle f(x) = 1+\frac{x}{1}+\frac{x^2}{2}+\frac{x^3}{3}+............+\frac{x^7}{7}.$$ Then Differentiate
both side w . r to $x\;,$ We get $$\displaystyle f'(x)=1+x+x^2+x^3+..........+x^6$$
Now for max. and Minimum Put $$f'(x) = 0\Rightarrow 1+x+x^2+x^3+x^4+x^5+x^6 = 0$$
We can write $f'(x)$ as $$\displaystyle \left(x^3+\frac{x^2}{2}\right)^2+\frac{3}{4}x^4+x^3+x^2+x+1$$
So $$\displaystyle f'(x) = \left(x^3+\frac{x^2}{2}\right)^2+\frac{1}{4}\left[3x^4+4x^3+4x^2+4x+4\right]$$
So $$\displaystyle f'(x) = = \left(x^3+\frac{x^2}{2}\right)^2+\frac{1}{4}\left[\left(\sqrt{2}x^2+\sqrt{2}x\right)^2+(x^2)^2+2(x+1)^2+2\right]>0\;\forall x\in \mathbb{R}$$
So $f'(x) = 0$ does not have any real roots. So Using $\bf{LMVT}$ $f(x) = 0$ has at most one root.
In fact $f(x) = 0$ has exactly one root bcz $f(x)$ is of odd degree polynomial and it
will Cross $\bf{X-}$ axis at least one time.
My question is can we solve it any other way, i. e without using Derivative test.
Help me , Thanks
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An odd polynomial has at least one real root is due to the fact that complex roots come in conjugate that is if $\alpha$ is a complex root of a polynomial then $\overline{\alpha}$ is a root too.
|
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.