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Computing $\zeta(6)=\sum\limits_{k=1}^\infty \frac1{k^6}$ with Fourier series. Let $ f$ be a function such that $ f\in C_{2\pi}^{0}(\mathbb{R},\mathbb{R}) $ (f is $2\pi$-periodic) such that $ \forall x \in [0,\pi]$: $$f(x)=x(\pi-x)$$
Computing the Fourier series of $f$ and using Parseval's identity, I have computed $\zeta(2)$ and $\zeta(4)$.
How can I compute $ \zeta(6) $ now?
Fourier series of $ f $:
$$ S(f)= \frac{\pi^2}{6}-\sum_{n=1}^{\infty} \frac{\cos(2nx)}{n^2}$$
$$ x=0, \zeta(2)=\pi^2/6$$
|
I have posted here in Portuguese a recursive method based on the computation of the Fourier trigonometric series expansion for the function defined in $\left[ -\pi ,\pi \right] $ by $f(x)=x^{2p}$ and extended to all of ${\mathbb R}$ periodically with period $2\pi.$ This is a shorter description than the original. In this reply I outline the case $\zeta(4)$. For $p=3$ the expansion is
$$x^{6}=\dfrac{\pi ^{6}}{7}+2\displaystyle\sum_{n\ge 1}^{}\left( \left( \dfrac{6}{n^{2}}\pi ^{4}-\dfrac{120}{n^{4}}\pi ^{2}+\dfrac{720 }{n^{6}}\right)\cos n\pi \right) \cos nx.\tag{1}$$
The computation is as follows:
$$\begin{equation*}
f(x)=x^{2p}=\frac{a_{0,2p}}{2}+\sum_{n=1}^{\infty }\left( a_{n,2p}\cos
nx+b_{n,2p}\sin nx\right) ,
\end{equation*}$$
where the coefficients are given by the following integrals
$$\begin{eqnarray*}
a_{0,2p} &=&\frac{1}{\pi }\int_{-\pi }^{\pi }x^{2p}\;\mathrm{d}x=\frac{2\pi ^{2p}}{2p+1},
\\
a_{n,2p} &=&\frac{1}{\pi }\int_{-\pi }^{\pi }x^{2p}\cos nx\;\mathrm{d}x=\frac{2}{\pi }
\int_{0}^{\pi }x^{2p}\cos nx\;\mathrm{d}x, \\
b_{n,2p} &=&\frac{1}{\pi }\int_{-\pi }^{\pi }x^{2p}\sin nx\;\mathrm{d}x=0.
\end{eqnarray*}$$
The series expansion is thus
$$\begin{equation*}
x^{2p}=\frac{\pi ^{2p}}{2p+1}+\frac{2}{\pi }\sum_{n=1}^{\infty }\cos
nx\int_{0}^{\pi }t^{2p}\cos nt\;\mathrm{d}t.\tag{2}
\end{equation*}$$
For $f(\pi )=\pi ^{2p}$ we obtain
$$
\begin{equation*}
\pi ^{2p}=\frac{\pi ^{2p}}{2p+1}+\frac{2}{\pi }\sum_{n=1}^{\infty }\cos n\pi
\int_{0}^{\pi }t^{2p}\cos nt\;\mathrm{d}t,
\end{equation*}$$
where the integral
$$
\begin{equation*}
I_{n,2p}:=\int_{0}^{\pi }x^{2p}\cos nx\;\mathrm{d}x
\end{equation*}$$
satisfies the following recurrence, as can be shown by integration by parts
$$\begin{equation*}
I_{n,2p}=\frac{2p}{n^{2}}\pi ^{2p-1}\cos n\pi -\frac{2p(2p-1)}{n^{2}}
I_{n,2\left( p-1\right) },\qquad I_{n,0}=0.\tag{3}
\end{equation*}$$
*
*For $p=1$, we get
$$\begin{equation*}
I_{n,2}=\frac{2}{n^{2}}\pi\cos n\pi.
\end{equation*}$$
and
$$\begin{eqnarray*}
\pi ^{2} &=&\frac{\pi ^{2}}{3}+\frac{2}{\pi }\sum_{n=1}^{\infty }\cos n\pi
\cdot I_{n,2} \\
&=&\frac{\pi ^{2}}{3}+\frac{2}{\pi }\sum_{n=1}^{\infty }\cos n\pi \left(
\frac{2}{n^{2}}\pi \cos n\pi \right) \\
&=&\frac{\pi ^{2}}{3}+4\sum_{n=1}^{\infty }\frac{1}{n^{2}} \\
&\Rightarrow &\zeta (2)=\sum_{n=1}^{\infty }\frac{1}{n^{2}}=\frac{\pi ^{2}}{6
}
\end{eqnarray*}$$
*For $p=2$, we get
$$
\begin{equation*}
I_{n,4}=\left( \frac{4\pi ^{3}}{n^{2}}-\frac{24\pi }{n^{4}}\right) \cos n\pi
\end{equation*}$$
and
$$
\begin{eqnarray*}
\pi ^{4} &=&\frac{\pi ^{4}}{5}+\frac{2}{\pi }\sum_{n=1}^{\infty }\cos n\pi
\cdot I_{n,4}=\frac{\pi ^{4}}{5}+\frac{4\pi ^{4}}{3}-48\sum_{n=1}^{\infty }
\frac{1}{n^{4}} \\
&\Rightarrow &\zeta (4)=\sum_{n=1}^{\infty }\frac{1}{n^{4}}=\frac{\pi ^{4}}{
90}.
\end{eqnarray*}$$
*Finally for $p=3$, we get
$$\begin{equation*}
I_{n,6}=\left( \frac{6\pi ^{5}}{n^{2}}-\frac{120\pi ^{3}}{n^{4}}+\frac{720}{
n^{6}}\right) \cos n\pi
\end{equation*}$$
and
$$
\begin{equation*}
\pi ^{6}=\frac{\pi ^{6}}{7}+2\sum_{n=1}^{\infty }\left( \frac{6\pi ^{4}}{
n^{2}}-\frac{120\pi ^{2}}{n^{4}}+\frac{720}{n^{6}}\right),
\end{equation*}$$
from which the result follows
$$\zeta(6)=
\begin{equation*}
\sum_{n=1}^{\infty }\frac{1}{n^{6}}=\frac{\pi ^{6}}{945}.
\end{equation*}$$
Plots of the periodic function defined in $\left[ -\pi ,\pi \right] $ by $f(x)=x^{6}$ (blue curve) and of the partial sum with the first 10 terms of its Fourier trigonometric series (red curve).
This method generates recursively the sequence $(\zeta(2p))_{p\ge 1}$.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/115981",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "47",
"answer_count": 6,
"answer_id": 2
}
|
Evaluate $\int_0^1 \frac{\ln(1+bx)}{1+x} dx $
What is $ \displaystyle\int_0^1 \frac{\ln(1+bx)}{1+x} dx $?
I call it $f(b)$ and differentiate with respect to be $b,$ a bit of partial fractions and the $x$ integral can be done. Then I integrate with respect to $b$ and get a bit lost.
Can some of the terms be expressed in terms of dilogarithms? I get lost in the details! Could we avoid all this just go straight to dilogarithms (with a cunning substitution)?
|
Let $x = \frac{1}{b} \frac{1-u}{u}$. The integration range $0\leqslant x \leqslant 1$ translates to $\frac{1}{1+b} \leqslant u \leqslant 1$, for $b>0$
$$
\frac{\log(1+ b x)}{1+x}\mathrm{d} x \to \frac{-\log(u)}{1 - (1-b) u} \frac{\mathrm{d} u}{u}
= \frac{\log(u)}{u} \left(1 + \frac{(1-b) u}{1-(1-b) u} \right)
$$
Now
$$ \begin{eqnarray}
\int \frac{\log(u)}{1 - (1-b) u} \frac{\mathrm{d} u}{u} &\stackrel{\color\red{\text{by parts}}}{=}& \frac{\log^2(u)}{2} - \log(u) \cdot \log(1-(1-b)u) - \int \frac{\log(1-(1-b)u)}{u} \mathrm{d} u \\
&=& \frac{\log^2(u)}{2} - \log(u) \cdot \log(1-(1-b)u) - \operatorname{Li_2}\left((1-b) u\right) + C
\end{eqnarray}
$$
where a definition of dilogarithm was used $\operatorname{Li_2}^\prime(z) = -\frac{\log(1-z)}{z}$. Now assuming $b$ is such that the anti-derivatives is smooth:
$$
\begin{eqnarray}
\int_0^1 \frac{\log(1+b x)}{1+x} \mathrm{d} x &=& \int\limits_{\frac{1}{1+b}}^1 \frac{-\log(u)}{1 - (1-b) u} \frac{\mathrm{d} u}{u} \\
&=& \operatorname{Li_2}\left(1-b\right) - \operatorname{Li_2}\left(\frac{1-b}{1+b}\right) + \frac{1}{2} \log^2\left(1+b\right) +\log\left(1+b\right) \log\left(\frac{2b}{1+b}\right)
\end{eqnarray}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/117924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 1
}
|
Prove: $\int_0^{\frac{\pi}{2}}t(\frac{\sin nt}{\sin t})^4dt<\frac{\pi^2n^2}{4}$ I have a question about integral. Prove:
$$\int_0^{\frac{\pi}{2}}t\left(\dfrac{\sin(nt)}{\sin(t)}\right)^4dt<\dfrac{\pi^2n^2}{4}$$
I have tried several methods including $\sin(t)\geq\frac{2t}{\pi}$, but I can't work it out.
|
Assume that $n\in \mathbb{N}$.
Then the ratio of sines is polynomial in cosines:
$$
\frac{\sin(n t)}{\sin(t)} = \cos((n-1) t) + \cos(t) \frac{\sin((n-1) t}{\sin(t)} = \ldots = \sum_{k=1}^n \cos((n-k) t) \cdot \cos^{k-1}(t) \leqslant n
$$
Since $\sin(t)$ is increasing on the interval $\left(0,\frac{\pi}{2}\right)$, and since $\frac{\sin(n t)}{\sin(t)} < n$ for $0<t<\frac{\pi}{2n}$, we have:
$$
\int_0^{\tfrac{\pi}{2}} t \cdot \left( \frac{\sin(n t)}{\sin(t)} \right)^4 \mathrm{d} t = \sum_{k=1}^n \int_{\tfrac{k-1}{n} \cdot \tfrac{\pi}{2}}^{\tfrac{k}{n} \cdot \tfrac{\pi}{2}} t \cdot \left( \frac{\sin(n t)}{\sin(t)} \right)^4 \mathrm{d} t < \\
\int_0^{\pi/(2n)} t n^4 \mathrm{d} t + \sum_{k=2}^n \int_{\tfrac{k-1}{n} \cdot \tfrac{\pi}{2}}^{\tfrac{k}{n} \cdot \tfrac{\pi}{2}} t \cdot \left( \frac{\sin(n t)}{\sin\left( \frac{k-1}{n} \cdot \tfrac{\pi}{2} \right)} \right)^4 \mathrm{d} t = \\
\frac{\pi^2 n^2}{8} + \sum_{k=2}^n \int_{\tfrac{k-1}{n} \cdot \tfrac{\pi}{2}}^{\tfrac{k}{n} \cdot \tfrac{\pi}{2}} t \cdot \left( \frac{\sin(n t)}{\sin\left( \frac{k-1}{n} \cdot \tfrac{\pi}{2} \right)} \right)^4 \mathrm{d} t
$$
The remaining bounding integral is not hard to compute:
$$
\int_{\tfrac{k-1}{n} \cdot \tfrac{\pi}{2}}^{\tfrac{k}{n} \cdot \tfrac{\pi}{2}} t \cdot \sin^4(n t) \mathrm{d} t \stackrel{t = \tfrac{(k-1)\pi}{2n} + \tfrac{u}{2}}{=} \int_0^{\tfrac{\pi}{2}} \frac{2u+ \pi(k-1)}{2n^2} \left( \frac{1+(-1)^k}{2} \cos^4 u + \frac{1-(-1)^k}{2} \sin^4 u \right) \mathrm{d} u = \frac{3 \pi^2 \cdot (2k-1) - 16 (-1)^k}{64 n^2}
$$
The upper bound then becomes:
$$
\frac{\pi^2 n^2}{8} + \sum_{k=2}^n \frac{3 \pi^2 \cdot (2k-1) - 16 (-1)^k}{64 n^2 \cdot \sin^4\left( \frac{k-1}{n} \cdot \tfrac{\pi}{2} \right)}
$$
It is easy to check numerically that this bound is more crude than the one you seek to establish.
Added
Notice the series expansion around $t=0$:
$$
\left(\frac{\sin(n t)}{\sin(t)} \right)^4 = n^4 \left( 1 - \frac{2}{3} (n^2-1) t^2 + \mathcal{o}(t^2) \right)
$$
This suggests looking for a bound in the form $\exp(-(n^2-1) t^2 \alpha)$. Suppose we fix a small enough $\alpha$, such that
$$
\forall_{0 < t < \tfrac{\pi}{2}} \left(\frac{\sin(n t)}{\sin(t)} \right)^4 \leqslant \exp(-(n^2-1) t^2 \alpha)
$$
Then
$$
\int_0^{\pi/2} t \cdot \left(\frac{\sin(n t)}{\sin(t)} \right)^4 \mathrm{d} t < \int_0^{\pi/2} t \exp(-(n^2-1) t^2 \alpha) \mathrm{d} t = n^4 \frac{1 - \exp(-\alpha (n^2-1) \pi^2/4)}{\alpha (n^2-1)} = \frac{ \pi^2 n^2}{8} \cdot \exp\left(-\frac{\pi^2 \alpha}{8} (n^2-1)\right) \cdot n^2 \operatorname{sinch}\left(\frac{\pi^2 \alpha}{8} (n^2-1)\right)
$$
where $\operatorname{sinch}(x) = \frac{\sinh(x)}{x}$ and is an increasing function of $x$, therefore:
$$
\frac{ \pi^2 n^2}{8} \cdot \exp\left(-\frac{\pi^2 \alpha}{8} (n^2-1)\right) \cdot n^2 \operatorname{sinch}\left(\frac{\pi^2 \alpha}{8} (n^2-1)\right) < \frac{ \pi^2 n^2}{8} \cdot \exp\left(-\frac{\pi^2 \alpha}{8} (n^2-1)\right) \cdot n^2 \operatorname{sinch}\left(\frac{\pi^2 \alpha}{8} n^2 \right) < \frac{ \pi^2 n^2}{8} \cdot \exp\left(\frac{\pi^2 \alpha}{8}\right)
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/120469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 0
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|
How I can find the value of $abc$ using the given equations? If I have been given the value of
$$\begin{align*}
a+b+c&= 1\\
a^2+b^2+c^2&=9\\
a^3+b^3+c^3 &= 1
\end{align*}$$
Using this I can get the value of
$$ab+bc+ca$$
How i can find the value of $abc$ using the given equations?
I just need a hint.
I have tried by squaring the equations.
But could not get it.
Thanks in advance.
|
In general, $$a^n + b^n + c^n = \sum_{i+2j+3k=n} (-1)^j \frac{n}{i+j+k}{i+j+k\choose i,j,k}s_1^is_2^js_3^k$$
where the sum is over non-negative $i,j,k$, and where $s_1=a+b+c$, $s_2=ab+ac+bc$ and $s_3=abc$.
In particular, when $n=3$ there are only three triples $(i,j,k)=(3,0,0),(1,1,0),(0,0,1)$, and you get:
$$a^3+b^3+c^3 = (a+b+c)^3 - 3(ab+ac+bc)(a+b+c) + 3abc$$
Now solve for $abc$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/120536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
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|
The solution set of the equation $|2x - 3| = - (2x - 3)$ The solution set of the equation $\left | 2x-3 \right | = -(2x-3)$ is
$A)$ {$0$ , $\frac{3}{2}$}
$B)$ The empty set
$C)$ (-$\infty$ , $\frac{3}{2}$]
$D)$ [$\frac{3}{2}$, $\infty$ )
$E)$ All real numbers
The correct answer is $C$
my solution:
$\ 2x-3 = -(2x-3)$ when $2x-3$ $\geqslant$ $0$ $\Rightarrow$ $x$ = $\frac{3}{2}$
$-(2x-3) = -(2x-3)$ when $2x-3$ $<$ $0$ $\Rightarrow$ $0$ = $0$
I can't get how the answer is presented in interval notation (-$\infty$ , $\frac{3}{2}$].
Any help is appreciated.
|
$|2x-3|=\begin{cases}
3-2x, & \text{if } x \leq \frac{3}{2} \\
2x-3, & \text{if } x > \frac{3}{2}
\end{cases}$
a) $|2x-3|=3-2x$ , hence :
$3-2x=-(2x-3)$
$0=0$ , therefore :
$S_a : x \in \left(-\infty, \frac{3}{2}\right]$
b) $|2x-3|=2x-3$ , hence :
$2x-3=-(2x-3)$
$4x=6$
$x=\frac{3}{2}$ , therefore :
$S_b : x \in \emptyset $
Finally :
$S= S_a \cup S_b \Rightarrow S : x \in \left(-\infty, \frac{3}{2}\right] $
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/121240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Let $n$ be a positive integer such that $\displaystyle{\frac{3+4+\cdots+3n}{5+6+\cdots+5n} = \frac{4}{11}}$ Let $n$ be a positive integer such that
$$\displaystyle{\frac{3+4+\cdots+3n}{5+6+\cdots+5n} = \frac{4}{11}}$$
then
$$\displaystyle{\frac{2+3+\cdots+2n}{4+5+\cdots+4n}} = \frac{m}{p}.$$
The question further "Is $m+p$ a prime?"
|
At first attempt, I was tempted to do $\displaystyle{3+4+\cdots+3n = \frac{3n(3n+1)}{2}-3}$, but there are $4$ expressions of that sort, it is better to find a general form of it like this
$$
\begin{align*}
k+(k+1)+(k+1)+\cdots+kn &= \frac{1}{2} \left[ kn(kn+1)-k(k-1) \right]\\
&= \frac{1}{2}\left[ k(n+1)(kn-k+1)\right] \tag{1}
\end{align*}
$$
Appying $(1)$ to $\displaystyle{\frac{3+4+\cdots+3n}{5+6+\cdots+5n}} = \frac{3(n+1)(3n-3+1)}{5(n+1)(5n-5+1)} =\frac{3(3n-1)}{5(5n-4)} = \frac{4}{11}$, leads us to $\displaystyle{\frac{3n-2}{5n-4}=\frac{20}{33}}$ and further to an expresion $99n-66=100n-80 \Rightarrow n=14$
$$
\displaystyle{\frac{2+3+\cdots+2n}{4+5+\cdots+4n}} =\frac{2(n+1)(2n-2+1)}{4(n+1)(4n-4+1)} = \frac{30\times27}{60\times53} =\frac{27}{106}
$$
$m+p=27+106=133$. But $133 = 7\times19$. Therefore the answer is "No, $m+p=133$ is not a prime".
(For those wondering what did I just change, for clarity I changed the right side to be $\frac{m}{p}$, the answer still stays)
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/121558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
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|
Finding a basis for the columnspace of a matrix
Find a linearly independent set of vectors that spans the same subspace of $R^4$ as that spanned by the vectors -
$$
\begin{bmatrix}
2 \\
-4 \\
-1 \\
-2 \\
\end{bmatrix}
,
\begin{bmatrix}
7 \\
-2 \\
7 \\
2 \\
\end{bmatrix}
,
\begin{bmatrix}
1 \\
2 \\
3 \\
2 \\
\end{bmatrix}
,
\begin{bmatrix}
3 \\
-2 \\
2 \\
0 \\
\end{bmatrix}
$$
So I start by reducing the matrix of these vectors to echelon form -
$$
\begin{bmatrix}
1 & -7 & -3 & -2 \\
0 & 21 & 7 & 7 \\
0 & -30 & -10 & -10 \\
0 & -12 & -4 & -4 \\
\end{bmatrix}
$$
$$
\begin{bmatrix}
1 & -7 & -3 & -2 \\
0 & 3 & 1 & 1 \\
0 & -3 & -1 & -1 \\
0 & -3 & -1 & -1 \\
\end{bmatrix}
$$
$$
\begin{bmatrix}
1 & -7 & -3 & -2 \\
0 & 3 & 1 & 1 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
\end{bmatrix}
$$
Then I get $x_2 = -\frac{1}{3}x_3 - \frac{1}{3}x_4$ and $x_1 = \frac{2}{3}x_3 - \frac{1}{3}x_4$
So this gives me the linearly independent vectors below that span the same subspaces as the original vectors -
$$
\begin{bmatrix}
1 \\
0 \\
\frac {2}{3} \\
-\frac {1}{3}\\
\end{bmatrix}
,
\begin{bmatrix}
0 \\
1 \\
-\frac {1}{3} \\
-\frac {1}{3} \\
\end{bmatrix}$$
But when I enter this answer I am told it is incorrect...anyone able to see where Im going wrong?
|
Start with a matrix whose columns are the vectors you have.
Then reduce this matrix to row-echelon form.
A basis for the columnspace of the original matrix is given by the columns in the original matrix that correspond to the pivots in the row-echelon form.
What you are doing does not really make sense because elementary row operations do not preserve the column space, but you are looking for a basis of the column space. In the end, what you are finding is a basis for the nullspace of the matrix, but what you are looking for is a basis for the columnspace.
Now, the nullspace and the rowspace are orthogonal to each other (in fact, orthogonal complements), so you could use your two vectors to find a basis for the column space, but it's too much extra work.
So: starting with the matrix of vectors you have,
$$\begin{align*}
\left(\begin{array}{rrrr}
2 & 7 & 1 & 3\\
-4 & -2 & 2 & -2\\
-1 & 7 & 3 & 2\\
-2 & 2 & 2 & 0
\end{array}\right) &\to
\left(\begin{array}{rrrr}
-1 & 7 & 3 & 2\\
-4 & -2 & 2 & -2\\
2 & 7 & 1 & 3\\
-2 & 2 & 2 & 0
\end{array}\right) \to
\left(\begin{array}{rrrr}
-1 & 7 & 3 & 2\\
0 & -30 & -10 & -10\\
0 & 21 & 7 & 7\\
0 & -12 & -4 & -4
\end{array}\right)\\
&\to \left(\begin{array}{rrrr}
-1 & 7 & 3 & 2\\
0 & 3 & 1 & 1\\
0 & 3 & 1 & 1\\
0 & 3 & 1 & 1
\end{array}\right) \to
\left(\begin{array}{rrrr}
\color{red}{-1} & 7 & 3 & 2\\
0 & \color{red}3 & 1 & 1\\
0 & 0 & 0 & 0\\
0 & 0 & 0 & 0
\end{array}\right)
\end{align*}$$
(Same thing you had)
Now, since the pivots of the row echelon form are in columns 1 and 2 (marked in red), that means that columns 1 and 2 of the original matrix are independent and span the column space. That is, you can take
$$\left(\begin{array}{r}
2 \\-4\\-1\\-2\end{array}\right)\quad\text{and}\quad
\left(\begin{array}{r}
7\\-2 \\ 7 \\ 2
\end{array}\right).$$
It should be easy to see that the two vectors are linearly independent. Also,
$$\left(\begin{array}{r}
1\\2\\3\\2\end{array}\right) = -\frac{2}{3}\left(\begin{array}{r}
2\\-4\\-1\\-2\end{array}\right) + \frac{1}{3}\left(\begin{array}{r}7\\-2\\7\\2\end{array}\right)$$
and
$$\left(\begin{array}{r}
3\\-2\\2\\0\end{array}\right) = \frac{1}{3}\left(\begin{array}{r}
2\\-4\\-1\\-2\end{array}\right) + \frac{1}{3}\left(\begin{array}{r}7\\-2\\7\\2\end{array}\right)$$
so the first two columns span the same space as all four.
The reason this works is that any linear combination between the columns is respected by elementary row operations (equivalently: elementary row operations do not change the nullspace). Since the columns that contain the pivots are necessarily a basis for the columnspace of the row-echelon form, the corresponding columns in the original form a basis. In fact, you can find the expressions I found above to show the other two columns are linear combinations by looking in the reduced-row echelon form instead of the original matrix; this is often easier.
This method extracts a subset of your original set that is a basis for the span of the set; that is, it gives you a way to pare your set down to a basis. If all you need is to find a basis, you can also begin by putting the vectors as the rows of a matrix and then performing elementary row operations until you get a matrix in row-echelon form. The (transpose of the) rows that contain the pivots in the row-echelon form give you a basis for the span of the original vectors. The reason this works is that elementary row operations don't change the row space, and the rows that contain the pivots are clearly a basis for the row space of the row-echelon form of the matrix. But you would not look for the nullspace in any case.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/121936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
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|
limit of $\lim_{x \to 0}\left ( \frac{1}{x^{2}}-\cot x\right )$ Help me with that problem, please.
$$\lim_{x \to 0}\left ( \frac{1}{x^{2}}-\cot x\right )$$
|
$$\lim\limits_{x \to 0} \left(\frac{1}{x^2} - \frac{1}{\tan x}\right) = \lim\limits_{x \to 0} -\left( \frac{x^4 \tan x - x^2 \tan^2 x}{x^4 \tan^2 x}\right) =
-\lim\limits_{x \to 0} \frac{(x^2\tan x)(x^2-\tan x)}{(x^2 \tan x)(x^2 \tan x)}$$
Cancelling out terms:
$$-\lim\limits_{x \to 0} \frac{x^2 - \tan x}{x^2 \tan x}$$
Apply L'Hopitals Rule
$$-\lim\limits_{x \to 0}\frac{x \cos2x + x - 1}{x(x+\sin 2x)} =-\frac{\lim\limits_{x \to 0}x + \lim\limits_{x \to 0}x \cos 2x - 1}{\lim\limits_{x \to 0}x(x+\sin 2x)} =-\frac{-1}{\lim\limits_{x \to 0}x(x+\sin2x)}$$
The limit of the products is the product of the limits.
$$\frac{1}{\lim\limits_{x \to 0}x(x+\sin2x)} = \frac{1}{(\lim\limits_{x \to 0}x)(\lim\limits_{x \to 0}(x + \sin 2x))}$$
Since $\lim\limits_{x \to 0}x = 0$,
$$\lim\limits_{x \to 0} = \left(\frac{1}{x^2} - \cot x\right) = \infty$$
|
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|
Evaluate and prove by induction: $\sum k{n\choose k},\sum \frac{1}{k(k+1)}$
*
*$\displaystyle
0\cdot \binom{n}{0} + 1\cdot \binom{n}{1} + 2\binom{n}{2}+\cdots+(n-1)\cdot \binom{n}{n-1}+n\cdot \binom{n}{n}$
*$\displaystyle\frac{1}{1\cdot 2} + \frac{1}{2\cdot 3}+\frac{1}{3\cdot 4} +\cdots+\frac{1}{(n-1)\cdot n}$
How do you find the sum of these and prove it by induction? Can someone help me get through this?
|
One way to do this:
1) You are looking for $\sum_{k=0}^n k\binom{n}{k}$.
First, look at $f(x)=(1+x)^n=\sum_{k=0}^n \binom{n}{k}x^k$, (by Newton's binomial theorem). Hence $2^n=\sum_{k=0}^n \binom{n}{k}$, by calculating $f(1)$. Now, define $g(x)=\sum_{k=0}^n k\binom{n}{k}x^k$. What you need is $g(1)$. Try expressing $g(x)$ through $f(x)$. (Hint: what is $g'(x)$?)
2) Notice that $\frac{1}{(k-1)k}=\frac{1}{k-1}-\frac{1}{k}$. Hence:
$\frac{1}{1\cdot 2} + \frac{1}{2\cdot 3}+\frac{1}{3\cdot 4} +\cdots+\frac{1}{(n-1)\cdot n}=\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\cdots+\left(\frac{1}{n-1}-\frac{1}{n}\right)$
|
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|
Compute $\sum_{i=1}^{2n} \frac{x^{2i}}{x^i-1}$ where $\{ x \in \mathbb{C}$ | $x^{2n+1} = 1, x \neq 1\}$ $\{ x \in \mathbb{C}$ | $x^{2n+1} = 1$ , $x \neq 1\}$
Compute $\displaystyle{\sum_{i=1}^{2n} \frac{x^{2i}}{x^i-1}}$
|
Since
$$(x^{\small{2n+1}}-1) = (x-1)(x^{\small{2n}}+x^{\small{2n-1}}+\cdots+x^{\small{2}}+x+1)=0$$
Therefore
$$(x^{\small{2n}}+x^{\small{2n-1}}+\cdots+x^{\small{2}}+x+1)=0 \hspace{4pt} ({\text{since}} \hspace{4pt} x \neq 1) \tag{1}$$
$$
\begin{align*}
S = \sum_{i=1}^{\small{2n}} \frac{x^{\small{2i}}}{x^i-1} &= \sum_{i=1}^{\small{2n}} x^i + \sum_{i=1}^{\small{2n}} \frac{x^{\small{i}}}{x^i-1}\\
&= -1 + \sum_{i=1}^{\small{2n}} \frac{x^{\small{i}}}{x^i-1} \hspace{12pt} {\text{from}} \hspace{4pt} (1) \\
&= -1 + \sum_{i=1}^{n} \frac{x^{\small{i}}}{x^i-1} + \sum_{i={\small{n+1}}}^{\small{2n}} \frac{x^{\small{i}}}{x^i-1} \tag{2}
\end{align*}
$$
Since $x^{\small{2n+1}} = 1$
$$ \frac{x^{\small{n+i}}}{x^{\small{n+i}}-1} = \frac{-1}{x^{\small{n+1-i}}-1} $$
Therefore $(2)$ can be simplified to
$$
\begin{align*}
& -1 + \sum_{i=1}^{n} \frac{x^{\small{i}}}{x^i-1} + \sum_{i={\small{n}}}^{\small{1}} \frac{-1}{x^{\small{n+1-i}}-1}\\
&= -1 + \sum_{i=1}^{n} \frac{x^{\small{i}}}{x^i-1} + \sum_{i={\small{1}}}^{\small{n}} \frac{-1}{x^{\small{i}}-1}\\
&= -1 + \sum_{i=1}^{n} \frac{x^{\small{i}}-1}{x^i-1} \\
&= n-1
\end{align*}
$$
|
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|
Limit: How to Conclude I have difficulty to conclude this limit ....; place of my attempts and results, can anyone help?
tanks in advance
$$\lim_{x\to +a}\, \left(1+6\left(\frac{\sin x}{x^2}\right)^x\cdot\frac{\log(1+10^x)}{x}\right),\quad a=+\infty,\,\,\,0,\,\,\,\,-\infty$$
1):$\,\,\,{a=+\infty}$
$$\lim_{x\to +\infty}\left(1+6\left(\frac{\sin x}{x^2}\right)^x\cdot\frac{\log(1+10^x)}{x}\right)$$
$$\sim\lim_{x\to +\infty}\, 1+6\left( \frac{\sin^x x}{x^{2x}} \cdot\frac{\log 10^x }{x}\right)=\lim_{x\to +\infty}\, 1+6\left( \frac{\sin^x x}{x^{2x}} \right)$$
$$\text{observing}\,\,\,\,\,\left|\frac{\sin x}{x^2} \right|<\frac{1}{x^2}, \text{so} \left|\left(\frac{\sin x}{x^2}\right)^x\right|<\frac{1}{x^{2x}}\to 0\,\,\,\, x\to+\infty:$$
$$\text{infact we have} \lim_{x\to+\infty} \frac{1}{x^{2x}}=\lim_{x\to +\infty} e^{-2x\log x}=0$$
$$=\lim_{x\to +\infty}\, 1+6\left( \frac{\sin^x x}{x^{2x}} \right)=1+6\cdot0=1
$$
2):$\,\,\,{a=0}$
$$\lim_{x\to 0}\, \left(1+6\left(\frac{\sin x}{x^2}\right)^x\cdot\frac{\log(1+10^x)}{x}\right) \sim\lim_{x\to 0}\, 1+\frac{6\log2}{x}\left(\frac{ x}{x^2}\right)^x$$
$$\sim\lim_{x\to0}1+\frac{6\log2}{x}\left(\frac{1}{x}\right)^x$$
$$=\lim_{x\to 0} 1+\frac{6\log2}{x}e^{ x\ln \left(\frac{1}{x}\right)}\to \lim_{x\to 0}\, x\ln \left(\frac{1}{x}\right)=\lim_{x\to 0}\, \frac{\ln \left(\frac{1}{x }\right)}{\frac{1}{x}}=0\to e^0=1$$
$$=\lim_{x\to 0}\,1+\frac{6\log2}{x}\cdot1=+\infty$$
3):$\,\,\,{a=-\infty}$
let $x=-t,\,\,\,$if$\,\,\,\ x\to -\infty\,\,\,$we have$ \,\,\,\ t\to+\infty,\,\,\,$ and so :
$$\lim_{t\to +\infty}\, \left(1+6\left(\frac{\sin(-t)}{t^2}\right)^{-t}\cdot\frac{\log(1+10^{-t})}{-t}\right)$$
$$=\lim_{t\to +\infty}\, \left(1-6\left(-\frac{\sin t }{t^2}\right)^{-t}\cdot\frac{\log\left(1+\frac{1}{10^t}\right)}{t}\right)$$
$$=1-6\cdot\lim_{t\to +\infty}\, \left[\left(-\frac{t^2}{\sin t }\right)^{t}\cdot\frac{\log\left(1+\frac{1}{10^t}\right)}{t}\right]$$
$$\stackrel{(\bf T)}{=}1-6\cdot\lim_{t\to +\infty}\, \left[\left(-\frac{t^2}{\sin t }\right)^{t}\cdot \left(\frac{1}{10^t}-\frac{1}{2\cdot10^{2t}}\cdot\frac{1}{t}\right) \right]$$
.... but here i'm lost ....
|
As pointed by David Mitra in comments, the function $$f(x) = \left\{1 + 6\left(\frac{\sin x}{x^{2}}\right)^{x}\frac{\log(1 + 10^{x})}{x}\right\}$$ is not well defined whenever $x \to \pm \infty$ because $\sin x$ becomes negative. The same holds when $x \to 0$. However if we restrict ourselves to $x \to 0^{+}$ then the function is well defined near $0$ and hence we may try to evaluate its limit as $x \to 0^{+}$. Clearly we can see that when $x \to 0^{+}$ then $\log(1 + 10^{x}) \to \log 2$ so that we ideally need to take care of the part $$g(x) = \frac{(\sin x)^{x}}{x^{2x + 1}}$$ which is better handled by taking logarithm. We have $$\begin{aligned}\log g(x) &= x\log\sin x - (2x + 1)\log x\\
&= x\log\left(\frac{\sin x}{x}\right) + x\log x - 2x\log x - \log x\\
&= x\log\left(\frac{\sin x}{x}\right) - x\log x - \log x\end{aligned}$$ Now as $x \to 0^{+}$ we can see that $(\sin x)/x \to 1$ so that first term tends to $0$. The second term $x\log x$ also tends to $0$ and the last term $-\log(x)$ tends to $\infty$. So that $\log g(x)$ tends to $\infty$ as $x \to 0^{+}$ and hence $g(x)$ also tends to $\infty$ as $x \to 0^{+}$. Since $f(x) = 1 + 6g(x)\log(1 + 10^{x})$ it follows that $f(x) \to \infty$ as $x \to 0^{+}$.
|
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|
Rearranging a formula, transpose for A2 - I'm lost Given the formula:
$$ q = A_1\sqrt\frac{2gh}{(\frac{A_1}{A_2})^2-1} $$
Transpose for $A_2$
I have tried this problem four times and got a different answer every time, none of which are the answer provided in the book. I would very much appreciate if someone could show me how to do this step by step.
The answer from the book is:
$$ A_2=\sqrt\frac{A_1^2q^2}{2A_1^2gh+q^2} $$
The closest I can get is the following:
$$ q = A_1\sqrt\frac{2gh}{(\frac{A_1}{A_2})^2-1} $$
$$ \frac{q^2}{A_1^2} = \frac{2gh}{(\frac{A_1}{A_2})^2-1} $$
Invert:
$$ \frac{A_1^2}{q^2} = \frac{(\frac{A_1}{A_2})^2-1}{2gh} $$
Multiply both sides by $2gh$:
$$ 2gh\frac{A_1^2}{q^2} = (\frac{A_1}{A_2})^2-1 $$
$$ \frac{2ghA_1^2}{q^2} = (\frac{A_1}{A_2})^2-1 $$
Add 1 to both sides and re-arrange:
$$ \frac{A_1^2}{A_2^2} = \frac{2ghA_1^2}{q^2} +1 $$
Invert again:
$$ \frac{A_2^2}{A_1^2} = \frac{q^2}{2ghA_1^2} +1 $$
Multiply by $A_1^2$:
$$ A_2^2 = \frac{A_1^2q^2}{2ghA_1^2} +1 $$
Get the square root:
$$ A_2 = \sqrt{\frac{A_1^2q^2}{2ghA_1^2}+1} $$
I cannot see where the $q^2$ on the bottom of the textbook answer comes from.
|
$$q = A_1\sqrt\frac{2gh}{(\frac{A_1}{A_2})^2-1}$$
$$q^2=(A_1)^2\frac{2gh}{(\frac{A_1}{A_2})^2-1}$$
$$(\frac{A_1}{A_2})^2-1=(A_1)^2\frac{2gh}{q^2}$$
$$(\frac{A_1}{A_2})^2=(A_1)^2\frac{2gh}{q^2}+1$$
$$\frac{A_1}{A_2}=\sqrt{(A_1)^2\frac{2gh}{q^2}+1}$$
$$A_2=\frac{A_1}{\sqrt{(A_1)^2\frac{2gh}{q^2}+1}}$$
|
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|
How do we prove $\cos(\pi/5) - \cos(2\pi/5) = 0.5$ ?. How do we prove $\cos(\pi/5) - \cos(2\pi/5) = 0.5$ without using a calculator.
Related question: how do we prove that $\cos(\pi/5)\cos(2\pi/5) = 0.25$, also without using a calculator
|
Note that: $\cos{2x} = \cos^{2}{x} - \sin^{2}{x} = 2\:\cos^{2}{x} - 1$. Therefore you have $\cos \frac{2\pi}{5} = 2\:\cos^{2}\frac{\pi}{5} - 1$
Now,
\begin{align*}
\cos\frac{\pi}{5} - \cos\frac{2\pi}{5} = \cos\frac{\pi}{5} - 2\: \cos^{2}\frac{\pi}{5}+1
\end{align*}
This is a quadratic equation of the form $2 x^{2} - x -1 =0$ and solving this will give you the value of $\cos\frac{\pi}{5}$ from which you can find the above value which you need.
|
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|
Integral with cube, square and zero starting point I don't know is there is a special rule or trick for this but I am trying to find
$$\int_{0}^1(x^3-3x^2) dx$$
I know that $\dfrac{1}{n}$ is the delta $x$ and this is where I do not know what to do next. I think that I want it to look something like
$$\lim \sum \dfrac{1}{n} \dfrac{k^{3}}{n} - n \dfrac{3k^{2}}{n}$$ But I am not sure what to do with coefficients yet.
I end up with something that looks like $\dfrac{3}{2} + 2n + \dfrac{2n}{6} + \dfrac{1}{2}$ which doesn't make sense since I have no many infinities.
|
So you have $\Delta x = \frac{1}{n}$.
And so$$\begin{align}
\int_0^1 f(x) dx &= \lim_{n\to \infty} \sum_{k=1}^{n} (\Delta x) f\left(0 + k\Delta x \right) \\
&= \lim_{n\to \infty} \sum_{k=1}^{n} \frac{1}{n} f\left(\frac{k}{n}\right) \\
&= \lim_{n\to \infty} \frac{1}{n}\sum_{k=1}^{n} \left(\frac{k}{n}\right)^3 - 3\left(\frac{k}{n}\right)^2 \\
&= \lim_{n\to \infty} \frac{1}{n}\sum_{k=1}^{n} \frac{1}{n^3}k^3 - \frac{3}{n^2}k^2 \\
&= \lim_{n\to \infty} \frac{1}{n}\left[\sum_{k=1}^{n} \frac{1}{n^3}k^3\right] - \frac{1}{n}\sum_{k=1}^{n}\frac{3}{n^2}k^2 \\
&= \lim_{n\to \infty} \frac{1}{n^4}\left[\sum_{k=1}^{n} k^3\right] - \frac{3}{n^3}\sum_{k=1}^{n}k^2 \\
&= ...
\end{align}
$$
And to find this you can use that
$$
\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6} \\
\sum_{k=1}^n k^3 = \left( \frac{n(n+1)}{2} \right)^2
$$
|
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|
How to prove $n^5 - n$ is divisible by $30$ without reduction How can I prove that prove $n^5 - n$ is divisible by $30$?
I took $n^5 - n$ and got $n(n-1)(n+1)(n^2+1)$
Now, $n(n-1)(n+1)$ is divisible by $6$.
Next I need to show that $n(n-1)(n+1)(n^2+1)$ is divisible by $5$.
My guess is using Fermat's little theorem but I don't know how.
|
$n^5-n=(n-1)n(n+1)(n^2+1)$. Rewrite $n^2+1$ as $5(n-1)+(n^2-5n+6)$ to obtain $n^5-n=5(n-1)^2n(n+1)+(n-3)(n-2)(n-1)n(n+1)$ and use the fact that $n!$ divides the product of n consecutive numbers.
|
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|
How was Euler able to create an infinite product for sinc by using its roots? In the Wikipedia page for the Basel problem, it says that Euler, in his proof, found that
$$\begin{align*}
\frac{\sin(x)}{x} &=
\left(1 - \frac{x}{\pi}\right)\left(1 + \frac{x}{\pi}\right)\left(1 - \frac{x}{2\pi}\right)\left(1 + \frac{x}{2\pi}\right)\left(1 - \frac{x}{3\pi}\right)\left(1 + \frac{x}{3\pi}\right) \cdots
\end{align*}$$
because the roots are at $\pm\pi, \pm2\pi, \pm3\pi, \cdots$ and finite polynomials are in this form (i.e. $(x-\text{root}_1)(x-\text{root}_2)\cdots$).
How was he able to do this? Why does this not simply make a polynomial function that has the roots same roots of $(\sin x)/x$? Can this method be used to make other trigonometric functions?
|
Nearly the same question was posted here recently. I hope this will add a little that is not in the other answers to this present question.
We know that $\dfrac{\sin x}{x}=0$ when $\sin x= 0$ and $x\neq0$, and we know that $\dfrac{\sin x}{x}$ "$=$" $1$ when $x=0$ (I think Euler's way of saying this is that $\sin x = x$ when $x$ is infinitely small). So this function should be $0$ when $x=\pm\pi$ or $\pm2\pi$ or $\pm3\pi$, etc., so it is
$$
\begin{align}
& \text{constant}\cdot(x-\pi)(x+\pi)(x-2\pi)(x+2\pi)(x-3\pi)(x+3\pi)\cdots \\[8pt]
& = \text{constant}\cdot(x^2-\pi^2)(x^2-4\pi^2)(x^2-9\pi^2)\cdots.
\end{align}
$$
When $x=0$, this is $(-\pi^2)(-4\pi^2)(-9\pi^2)\cdots$. But we saw above that when $x=0$, this is $1$. Hence we have
$$
\begin{align}
\frac{\sin x}{x} & = \frac{(x^2-\pi^2)(x^2-4\pi^2)(x^2-9\pi^2)\cdots}{(-\pi^2)(-4\pi^2)(-9\pi^2)\cdots} \\[8pt]
& = \left(1-\frac{x^2}{\pi^2}\right)\left(1-\frac{x^2}{4\pi^2}\right)\left(1-\frac{x^2}{9\pi^2}\right)\cdots.
\end{align}
$$
|
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|
$\int_0^{2\pi}\sin\frac{x}{2}\cos^2\frac{x}{2}\,dx$ $$\int_0^{2\pi}\sin\frac{x}{2}\cos^2\frac{x}{2}\,dx$$
Tried substitution ($u = \cos\frac{x}{2}$), but I get $-\frac{\cos^3\frac{x}{2}}{3}$ ($-\frac{2}{3}$) instead of the correct answer, which is $1\frac{1}{3}$
|
It makes it a little easier to do this simple substitution for you to not mess up with constant multiple like you did (You got an extra $\frac{1}{3}$)
Substitute first $\frac{x}{2}=t$. To see the limits, when $x=2\pi, t=\pi$ and when $x=0, u=0$ and $\mathrm{d}x = 2\mathrm{d}t$
$$
\begin{align*}
\int_0^{2\pi}\sin\frac{x}{2}\cos^2\frac{x}{2} \mathrm{d}x &= 2 \int_0^{\pi}\sin t \hspace{3pt}\cos^2 t \hspace{3pt} \mathrm{d}t\\
&= 2 \left|\frac{-\cos^3 t}{3} \right|_0^{\pi}\\
&= 2 \left[\left(\frac{1}{3}\right) - \left( \frac{-1}{3} \right) \right]\\
&= 2 \left(\frac{2}{3}\right) = \frac{4}{3}
\end{align*}
$$
|
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|
Definition of derivative $f(x) = \sqrt{3-5x}$ I am not sure how to factor this out
$$f(x) = \sqrt{3-5x}$$
I then make it $f(x) = \frac {\sqrt{3-5(x+h)} - \sqrt{3-5x}}{h}$
I tried to multiply by the first time + the second term from the numerator which I called x and y
$$\frac {x - y}{h} \cdot \frac {x + y}{x+y}$$
which gives me
$$\frac {x^2 - y^2}{h(x+y)}$$
From here it gets very difficult
$$\frac {5}{ h \sqrt{3-5(x+h)} - \sqrt{3-5x}}$$
|
$\displaystyle f(x) = \sqrt{3-5x}$. Hence, $\displaystyle f(x+h) = \sqrt{3 - 5(x+h)}$. Therefore, we get that $$f(x+h)-f(x) = \sqrt{3 - 5(x+h)} - \sqrt{3 - 5x} = \frac{(3 - 5(x+h)) - (3 - 5x)}{\sqrt{3 - 5(x+h)} + \sqrt{3 - 5x}} = \frac{-5h}{\sqrt{3 - 5(x+h)} + \sqrt{3 - 5x}}$$
Hence, $$\frac{f(x+h)-f(x)}{h} = - \frac{5}{\sqrt{3 - 5(x+h)} + \sqrt{3 - 5x}}$$
Now take the limit as $h \rightarrow 0$, to get $$f'(x) = \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} = - \lim_{h \rightarrow 0} \frac{5}{\sqrt{3 - 5(x+h)} + \sqrt{3 - 5x}} = -\frac{5}{\lim_{h \rightarrow 0} \sqrt{3 - 5(x+h)} + \sqrt{3 - 5x}} = - \frac{5}{2\sqrt{3 - 5x}}$$
|
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|
Help with Cardano's Formula I'm trying to understand how to solve cubic equations using Cardano's formula. To test the method, I expand $(x-3)(x+1)(x+2)=x^3-7x-6$. My hope is that the formula will produce the roots $-1,-2,3$. But the formula seems to make a mess of things: I compute that $\frac{q^2}{4}+\frac{p^3}{27}=\frac{100}{27}$, and so the formula gives me the baffling
\begin{equation}
\sqrt[3]{3+i \sqrt{\frac{100}{27}}}+\sqrt[3]{3-i \sqrt{\frac{100}{27}}}.
\end{equation}
I'd like to know if there is a straightforward way one or all of the roots $-1,-2,3$ from this expression. I've asked several people this question, and the usual punchline is that I've produced a proof that this expression is $-1,-2$ or $3$ (depending on the choice of cube root etc.) That is not my goal.
I found a book of Cardano's writings in the library, but it seems some of his writings have been lost. I'm convinced that he and his cohort had some method for doing this. So, does anyone know how to use the cubic formula for real? Specifically, in such a way as to recognize the output as a particular integer/rational number when it is one?
Thanks!
|
I know it has been a lot of time since you asked the question, but it is reasonable to assume that no one has answered you. Ok, here we go. When trying to reduce radicals in general, you try to find values for a, b $\in \mathbb{Q}$ such that $$\sqrt[n]{A \pm B\sqrt[m]{C}} = a \pm b\sqrt[m]{C}.$$Notice that $\sqrt[m]{C}$ stays the same. What we are going to change the original expression to such form: $$\sqrt[3]{3 \pm i \sqrt{\frac{100}{27}}} = \sqrt[3]{3 \pm \frac{10}{3}\frac{i}{\sqrt{3}}} = \sqrt[3]{3 \pm \frac{10}{9}(i\sqrt{3})}$$ We will now try to find values for a, b $\in \mathbb{Q}$ such that $$\sqrt[3]{3 \pm \frac{10}{9}(i\sqrt{3})} = a \pm b(i\sqrt{3}).$$Cubing both sides:$$3 \pm \frac{10}{9}(i\sqrt{3}) = (a^3 - 9ab^2) \pm (3a^2b - 3b^3)(i\sqrt{3})$$ Since a and b are rational numbers, then the previous equation is only true if:$$\begin{cases}a^3 - 9ab^2 = 3\\3a^2b - 3b^3 = \frac{10}{9}\end{cases}\Rightarrow\begin{cases}a^3 - 9ab^2 = 3\\27a^2b - 27b^3 = 10\end{cases}$$To solve that system of equations we are going to multiply one equation by the other:$$(3)(27a^2b - 27b^3) = (10)(a^3 - 9ab^2)\\81a^2b - 81b^3 = 10a^3 - 90 ab^2\\10a^3 - 81a^2b - 90ab^2 + 81b^3 = 0$$Dividing both sides by $b^3$:$$10(\frac ab)^3 - 81(\frac ab)^2 - 90(\frac ab) + 81 = 0$$We are now going to define x = a/b. Now, since a and b are rational numbers, x is also rational. Using the rational root theorem, we can find the solutions for our cubic equation:$$10x^3 - 81x^2 - 90x + 81 = 0\\x = -{3 \over 2} \lor x = {3 \over 5} \lor x = 9$$If x = -3/2, then a = -1 and b = 2/3.
If x = 3/5, then a = -1/2 and b = -5/6.
If x = 9, then a = 3/2 and b = 1/6.
Therefore:$$\sqrt[3]{3 \pm i \sqrt{\frac{100}{27}}} = -1 \pm i\frac{2\sqrt{3}}{3}\\\sqrt[3]{3 \pm i \sqrt{\frac{100}{27}}} = -\frac{1}{2} \mp i\frac{5\sqrt{3}}{6}\\\sqrt[3]{3 \pm i \sqrt{\frac{100}{27}}} = \frac{3}{2} \pm i\frac{\sqrt{3}}{6}$$And there you have it. I learned this method here, by the way, after comming across the same problem you did.
|
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|
In what case this equation has a solution(s)? I am trying to solve this problem saying, for what values of $a$ and $b$ (in which they are not zero), the equation $a(1-\sin(x))=b(1-\cos(x))$ has any solution(s)? Thank you very much. :)
|
$1- \sin(x) = (\cos(x/2) - \sin(x/2))^2$ and $1 - \cos(x) = 2 \sin^2(x/2)$. This gives us that $$\cot(x/2) - 1 = \pm \sqrt{2b/a}$$ i.e. $$\tan(x/2) = \frac1{1 \pm \sqrt{2b/a}}$$ i.e. $$x = 2 \arctan \left(\frac1{1 \pm \sqrt{2b/a}} \right)$$
Another way to approach is as follows. We have $$a \sin(x) - b \cos(x) = b-a$$ i.e. $$\frac{a \sin(x) - b \cos(x)}{\sqrt{a^2+b^2}} = \frac{b-a}{\sqrt{a^2+b^2}}$$ i.e. $$\sin(x + \phi) = \frac{b-a}{\sqrt{a^2+b^2}}$$
Hence, $$x = \arcsin \left(\frac{b-a}{\sqrt{a^2+b^2}} \right) - \phi$$ where $\phi$ is such that $\displaystyle \cos(\phi) = \frac{a}{\sqrt{a^2+b^2}}$ and $\displaystyle \sin(\phi) = -\frac{b}{\sqrt{a^2+b^2}}$. For $x \in \mathbb{R}$, we need $$-1 \leq \frac{b-a}{\sqrt{a^2+b^2}} \leq 1$$ i.e. $$\frac{(b-a)^2}{a^2+b^2} \leq 1$$ i.e. $$-2ab \leq 0$$ i.e. $$ab \geq 0$$ Hence, $a$ and $b$ must both be of the same sign.
|
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|
Evaluating $\int_{-20}^{20}\sqrt{2+t^2}\,dt$ I have this integral:
$$\int_{-20}^{20}\sqrt{2+t^2}\,dt$$
I tried solving it many times but without success.
The end result is this:
$$2\left( 10\sqrt{402}+\mathop{\mathrm{arcsinh}}(10\sqrt{2})\right).$$
I can't seem to get this end result. I got a few wrong ones but cant find this one. Perhaps it's wrong? Could anyone confirm it?
I tried Sage, and it calculates it correctly, but not with steps.
|
Here are two more ways to calculate $\int \sqrt{a^2+x^2}\ dx$.
*
*Let us make the substitution $\sqrt{a^2+x^2}=t-x$. After squaring, $a^2=t^2-2xt$, from which
$$
x=\frac{t^2-a^2}{2t}\,.
$$
Using the last expression, one finds
$$
\sqrt{a^2+x^2}=t-x=\frac{t^2+a^2}{2t}\,,
$$
and
$$
dx=\frac{t^2+a^2}{2t^2}dt.
$$
Plugging everything into the integral,
$$
\frac 14\int\left(\frac{2a^2}{t}+t+\frac{a^4}{t^3}\right)dt=\frac{a^2}{2}\ln t+\frac{t^4-a^4}{8t^2}+C.
$$
Note, that from the above $x\sqrt{a^2+x^2}=\frac{t^4-a^4}{4t^2}$, therefore we obtain finally
$$
\int\sqrt{a^2+x^2}\,dx=\frac{a^2}{2}\ln(x+\sqrt{a^2+x^2})+\frac{1}{2}x\sqrt{a^2+x^2}+C.
$$
*Consider the integration by parts:
$$
u=\sqrt{a^2+x^2},\quad v=x,\quad du=\frac{x}{\sqrt{x^2+a^2}}.
$$
Therefore,
$$
\int\sqrt{x^2+a^2}\,dx=x\sqrt{a^2+x^2}-\int\frac{x^2dx}{\sqrt{a^2+x^2}}=\\
x\sqrt{a^2+x^2}-\int\frac{x^2+a^2-a^2 dx}{\sqrt{a^2+x^2}}=\\x\sqrt{a^2+x^2}-\int\sqrt{a^2+x^2}\,dx+\int\frac{a^2dx}{\sqrt{a^2+x^2}}.
$$
It is known (and can easily be found with the variable change from my point 1) that
$$
\int\frac{dx}{\sqrt{a^2+x^2}}=\ln(x+\sqrt{a^2+x^2})+C,
$$
hence, finally,
$$
\int\sqrt{a^2+x^2}\,dx=\frac{a^2}{2}\ln(x+\sqrt{a^2+x^2})+\frac{1}{2}x\sqrt{a^2+x^2}+C.
$$
|
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|
Summation of Infinite Series $\sum_{n=1}^{\infty} \frac{1}{n 2^{2n+1}}$ Show That :
$$\sum_{n=1}^{\infty} \frac{1}{n 2^{2n+1}} = \ln \left(\frac{2}{\sqrt{3}}\right)$$
I could show convergence. (I dont need to show that this converges). However I couldn't figure how to show the value.
|
$$
\begin{align*}
\sum_{n=1}^{\infty} \frac{y^n}{n} &= -\ln \left(1-y\right) \hspace{15pt} {\textit{apply }} \hspace{5pt} y=\frac{1}{x^2}\\
\sum_{n=1}^{\infty} \frac{1}{n x^{2n}} &= -\ln \left(1-\frac{1}{x^2}\right) \\
\sum_{n=1}^{\infty} \frac{1}{n 2^{2n+1}} &= \frac{1}{2}\sum_{n=1}^{\infty} \frac{1}{n 2^{2n}} = -\frac{1}{2}\ln \left(1-\frac{1}{4}\right)\\
\end{align*}
$$
Do the rest to simplify and get what you want.
|
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|
Finding square roots of $\sqrt 3 +3i$ I was reading an example, where it is calculating the square roots of $\sqrt 3 +3i$.
$w=\sqrt 3 +3i=2\sqrt 3\left(\frac{1}{2}+\frac{1}{2}\sqrt3i\right)\\=2\sqrt 3(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3})$
Let $z^2=w \Rightarrow r^2(\cos(2\theta)+i\sin(2\theta))=2\sqrt 3(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3})$.
But how did they get from $\sqrt 3 +3i=2\sqrt 3\left(\frac{1}{2}+\frac{1}{2}\sqrt3i\right)=2\sqrt 3(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3})$?
And can one just 'let $z^2=w$' as above?
Edit:
$w=2\sqrt 3(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3})=z^2\\ \Rightarrow z=\sqrt{2\sqrt 3}(\cos\frac{\pi}{6}+i\sin\frac{\pi}{6})\\ \Rightarrow \sqrt{2\sqrt 3}\frac{\sqrt 3}{2} +i \sqrt{2\sqrt 3} \frac{1}{2}$
|
Wikipedia page on polar form of complex numbers is quite good.
Given a complex number $z = a + i b$, its absolute value $|z| = \sqrt{a^2+b^2}$, naturally the quotient $\frac{z}{|z|}$ has unit absolutely value, hence $\frac{z}{|z|} = \mathrm{e}^{i \theta} = \cos(\theta) + i \sin(\theta)$ for some angle $\theta$.
In the case at $a=\sqrt{3}$ and $b=3$, thus $\sqrt{a^2+b^2} = \sqrt{3+3^2} = 2 \sqrt{3}$. Therefore $\frac{z}{|z|} = \frac{\sqrt{3}}{2 \sqrt{3}} + i \frac{3}{2 \sqrt{3}} = \frac{1}{2} + i \frac{\sqrt{3}}{2}$. Solving for $\cos(\theta) = \frac{1}{2}$ and $\sin(\theta) = \frac{\sqrt{3}}{2}$ for $0 \leqslant \theta < 2\pi$ gives $\theta = \frac{\pi}{3}$.
Finding the square root proceeds as follows. Let $w = |w| \mathrm{e}^{i \phi}$, then
$$
|w|^2 \mathrm{e}^{2 i \phi} = 2 \sqrt{3} \mathrm{e}^{i \pi/3}
$$
Taking the absolute value we must have $|w|^2 = 2 \sqrt{3}$, hence $|w| = \sqrt{2} 3^{1/4}$. When solving for the angle $\phi$, remember that there are two roots for $0 \leqslant \phi <2 \pi$.
|
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How to solve this equation, when the unknown variable just disappears? This problem
$\sqrt{1-x^2} + \sqrt{3+x^2} = 2$
has the solution $x = 1$ and $x = -1$.
However, I always get stuck like this:
*
*$1-x^2 + 3+x^2 = 4$
*$4 = 4$
How do I isolate that darn unknown?
|
$(a+b)^2\neq a^2+b^2$ in general. In your case $\sqrt{1-x^2}+\sqrt{3+x^2}=2$ leads to $1-x^2+2\sqrt{1-x^2}\sqrt{3+x^2}+3+x^2=4$, which gives you $2\sqrt{1-x^2}\sqrt{3+x^2}=0$. Since $\sqrt{3+x^2}\neq0$ for all real $x$, so $\sqrt{1-x^2}=0$. Hence $x=\pm1$.
|
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|
limit of power of fraction of sums of sines Find the following limit:
$$\lim_{n\to\infty} \left(\frac{{\sin\frac{2}{2n}+\sin\frac{4}{2n}+\cdot \cdot \cdot+\sin\frac{2n}{2n}}}{{\sin\frac{1}{2n}+\sin\frac{3}{2n}+\cdot \cdot \cdot+\sin\frac{2n-1}{2n}}}\right)^{n}$$
I thought of some $\sin(x)$ approximation formula, but it doesn't seem to work.
|
Let $f : [0, 1] \to [0, \infty)$ be of the class $C^1$ and not identically zero. Then by Mean Value Theorem, we have
$$ \sum_{k=1}^{n} f \left( \tfrac{2k}{2n} \right) = \sum_{k=1}^{n} \left( f \left( \tfrac{2k-1}{2n} \right) + f' (x_{n,k}) \frac{1}{2n} \right) $$
for some $x_{n,k} \in \left(\frac{2k-1}{2n}, \frac{2k}{2n} \right)$. Letting
$$ I_n = \frac{1}{n} \sum_{k=1}^{n} f \left( \tfrac{2k-1}{2n} \right) \quad \text{and} \quad J_n = \frac{1}{n} \sum_{k=1}^{n}f' (x_{n,k}),$$
We have
$$I_n \to I := \int_{0}^{1} f(x) \; dx \quad \text{and} \quad J_n \to J := \int_{0}^{1} f'(x) \; dx.$$
Therefore we obtain
$$ \left[ \frac{\sum_{k=1}^{n} f \left( \frac{2k}{2n} \right)}{\sum_{k=1}^{n} f \left( \frac{2k-1}{2n} \right)} \right]^{n} = \left( \frac{nI_n + \frac{1}{2}J_n}{n I_n} \right)^{n} = \left( 1 + \frac{1}{n}\frac{J_n}{2I_n} \right)^{n} \xrightarrow[n\to\infty]{} \exp \left( \frac{J}{2I} \right). $$
Now plugging $f(x) = \sin x$, the corresponding limit is $\exp \left( \frac{1}{2} \cot \frac{1}{2} \right)$.
|
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|
Linear Congruences I understand how to solve an equation like:
$$17x \equiv 3 \pmod 5$$
But is there a known method for solving a congruence of the form:
$$17 \equiv 3\frac{\sqrt{x}}{2} \pmod 5\quad ?$$
Is it as simple as moving the $x$ through operations and applying the congruence to the new equation?
$$(34/3)-\sqrt{x}\equiv 0 \pmod5$$
or
$$(14)/(\sqrt{x})\equiv3\pmod5$$
Or is the position of $x$ insignificant?
|
You have to be careful with square roots in modular equations, since not every number has a square root (much like you would for integers), and when they do, there is no way to select one the way we do with real number (where we always pick the nonnegative one). That is, $\sqrt{x}$ is not well-defined modulo $5$. For example, if $x=4$, does $\sqrt{x}$ represent $2$, or does it represent $3$? Both numbers, when squared, are congruent to $4$ modulo $5$.
But you can work essentially the same way as you do with equations. From
$$17\equiv \frac{3\sqrt{x}}{2}\pmod{5}$$
we note that $17\equiv 2\pmod{5}$; multiplying through by $2$ we get
$$4\equiv 3\sqrt{x}\pmod{5}.$$
The multiplicative inverse of $3$ modulo $5$ is $2$, so multiplying through by $2$ and noting that $8\equiv 3\pmod{5}$ we get
$$3\equiv \sqrt{x}\pmod{5}.$$
So $9\equiv x \pmod{5}$. Hence $x\equiv 4\pmod{5}$. Note that $4$ does have a square root modulo $5$; in fact, it has two square roots: $2$ and $3$. However, $\sqrt{x}=2$ does not satisfy the original equation, so you need to have $x\equiv 4\pmod{4}$ and $\sqrt{x}\equiv 3\pmod{5}$.
|
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|
Elementary Real Analysis - Let me know if I'm on the right track So I'm guessing this is a pretty simple example for this topic but I just want to check myself as I'm new to this analysis area and not sure that what I'm saying is mathematically sound..
The question is show that $\lim\limits_{x \to -\infty}\frac{x^2+1}{x^2-1} =1$.
So we must show as $x \to -\infty$, $\left|\tfrac{x^2+1}{x^2-1} -1\right| < \epsilon$.
My attempt is :
\begin{align*}
\left|\dfrac{x^2+1}{x^2-1} -1\right| &=\left|\dfrac{x^2+1}{x^2-1} - \dfrac{x^2-1}{x^2-1} \right| \\\\
&=\left|\dfrac{(x^2+1)-(x^2-1)}{x^2-1} \right| \\\\
&=\left|\dfrac{x^2-x^2+1+1}{x^2-1}\right| \\\\
&=\left|\dfrac{2}{x^2-1}\right| < \left|\dfrac{2}{x^2-4}\right| = \left|\dfrac{2}{(x+2)(x-2)}\right| \\
\end{align*}
Now is where I'm not 100 percent sure that what I'm doing is right, can we then say that as $x$ approaches negative infinity, $(x+2)$ and $(x-2)$ become very large and negative, and therefore $2/(x+2)(x-2)$ becomes smaller and smaller and so, for any $\epsilon>0$,
$$\epsilon > \left|\dfrac{2}{(x+2)(x-2)}\right| > \left|\dfrac{2}{x^2-1}\right| = \left|\dfrac{x^2+1}{x^2-1} -1\right|$$ thus proving the original problem ... is this ok/rigorous?
please dont be too hard on me I'm really still just trying to grasp the ideas and understand exactly what we are 'allowed' to do .. thanks for any help
|
Assume $\epsilon>0$, then
$$\begin{align*} \left|\frac{x^2+1}{x^2-1} - 1\right| < \epsilon
&\Rightarrow \left|\frac{2}{x^2-1}\right| < \epsilon \\
&\Rightarrow \left|\frac{x^2-1}{2}\right| > 1/\epsilon \\
&\Rightarrow |x^2-1| > \frac{2}{\epsilon}. \end{align*}
$$
when $x^2-1 > 0$, $x^2>1 $, $|x| > 1$,
so, $(x^2-1)>2/\epsilon$ .... (when $|x|>1$)
$x^2 > 2/\epsilon+1$
$|x| > \sqrt{2/\epsilon+1}$
therefore :
Given any $\epsilon > 0 $, $\left|\frac{x^2+1}{x^2-1} - 1\right| < \epsilon$
when $|x| > \sqrt{2/ϵ+1}$.
ta da !
references : M Turgeon that king !
|
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|
How can we produce another geek clock with a different pair of numbers? So I found this geek clock and I think that it's pretty cool.
I'm just wondering if it is possible to achieve the same but with another number.
So here is the problem:
We want to find a number $n \in \mathbb{Z}$ that will be used exactly $k \in \mathbb{N}^+$ times in any mathematical expresion to produce results in range $[1, 12]$. No rounding, is allowed, but anything fancy it's ok.
If you're answering with an example then use one pair per answer.
I just want to see that clock with another pair of numbers :)
Notes for the current clock:
1 o'clock: using 9 only twice, but it's easy to use it 3 times with many different ways. See comments.
5 o'clock: should be $\sqrt{9}! - \frac{9}{9} = 5$
|
For $n=4$ and $k=5$ here is a solution:
$\frac{4}{\left(4+\left(4 \times \left(4-4\right)\right)\right)}=1$
$\left(4-\left(4 \times \frac{4}{\left(4+4\right)}\right)\right)=2$
$\left(4+\frac{4}{\left(4-\left(4+4\right)\right)}\right)=3$
$\left(4+\left(4+\left(4-\left(4+4\right)\right)\right)\right)=4$
$\left(4-\frac{4}{\left(4-\left(4+4\right)\right)}\right)=5$
$\left(4+\left(4 \times \frac{4}{\left(4+4\right)}\right)\right)=6$
$\frac{4}{\left(4 \times \frac{4}{\left(4+4!\right)}\right)}=7$
$\left(4 \times \left(4 \times \frac{4}{\left(4+4\right)}\right)\right)=8$
$\left(4-\left(\frac{4}{4}-\frac{4!}{4}\right)\right)=9$
$\left(4+\frac{4}{\left(4 \times \frac{4}{4!}\right)}\right)=10$
$\frac{4}{\left(4 \times \frac{4}{44}\right)}=11$
$\left(4-\left(4-\left(4+\left(4+4\right)\right)\right)\right)=12$
|
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|
Sequence convergence and parentheses insertion find an example for a series $a_{n}$
that satisfies the following:
*
*$a_{n}\xrightarrow[n\to\infty]{}0$
*${\displaystyle \sum_{n=1}^{\infty}a_{n}}$
does not converges
*There is a way to insert parentheses so ${\displaystyle \sum_{n=1}^{\infty}a_{n}}$
will converges.
I was thinking about the series:$ 1-1+\frac{1}{2}+\frac{1}{2}-\frac{1}{2}-\frac{1}{2}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}-\frac{1}{4}-\frac{1}{4}-\frac{1}{4}-\frac{1}{4}+...$
But I don't know how to prove 2.
Also will be nice to hear another examples, if any.
|
The series $$1-1+\frac{1}{2}+\frac{1}{2}-\frac{1}{2}-\frac{1}{2}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}-\frac{1}{4}-\frac{1}{4}-\frac{1}{4}-\frac{1}{4}+...$$
is indeed divergent, because it has infinitely many partial sums equal to $1$, and infinitely many partial sums equal to $0$. Inserting parentheses as
$$\left(1-1\right)+\left(\frac{1}{2}+\frac{1}{2}-\frac{1}{2}-\frac{1}{2}\right)+\left(\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}-\frac{1}{4}-\frac{1}{4}-\frac{1}{4}-\frac{1}{4}\right)+...$$
yields a convergent series.
More generally: take any divergent series in which $\limsup S_N\ge 0$ and $\liminf S_N\le 0$ ($S_N$ are partial sums). Split every term into several, to ensure $a_n\to 0$. Insert $k$th closing parenthesis when the partial sum drops below $1/k$ in absolute value.
|
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|
simplify $ (-2 + 2\sqrt3i)^{\frac{3}{2}} $? How can I simplify $ (-2 + 2\sqrt3i)^{\frac{3}{2}} $ to rectangular form $z = a+bi$?
(Note: Wolfram Alpha says the answer is $z=-8$. My professor says the answer is $z=\pm8$.)
I've tried to figure this out for a couple hours now, but I'm getting nowhere.
Any help is much appreciated!
|
$$(-2 + 2\sqrt3i) = 4 \exp\left(\frac{2\pi}{3}i\right) = 4 \cos \left(\frac{2\pi}{3}\right) + 4 \sin \left(\frac{2\pi}{3}\right) i$$
and I would say $$\left(4 \exp\left(\frac{2\pi}{3}i\right)\right)^{\frac{3}{2}} = 4^{\frac{3}{2}} \exp\left(\frac{3}{2} \times \frac{2\pi}{3}i\right) =8 \exp(\pi i) = -8. $$
I think using $(-2 + 2\sqrt3i) = 4 \exp\left(\frac{8\pi}{3}i\right)$ or $4 \exp\left(-\frac{4\pi}{3}i\right)$ here would be unconventional.
To get an answer of $\pm 8$ you would need to believe $\sqrt{-2 + 2\sqrt3i} = -1-\sqrt3 i$ as well as $1+\sqrt3 i$ and while the square of each of them gives the intended value, I would take what I regard as the principal root giving a single answer, and so does Wolfram Alpha.
It is like saying $\sqrt{4} = 2$ alone even though $(-2)^2=4$ too.
|
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Calculating the shortest possible distance between points Question:
Given the points $A(3,3)$, $B(0,1)$ and $C(x,0)$ where $0 < x < 3$, $AC$ is the distance between $A$ and $C$ and $BC$ is the distance between $B$ and $C$. What is x for the distance $AC + BC$ to be minimal?
What have I done?
I defined the function $AC + BC$ as:
$\mathrm{f}\left( x\right) =\sqrt{{1}^{2}+{x}^{2}}+\sqrt{{3}^{2}+{\left( 3-x\right) }^{2}}$
And the first derivative:
$\mathrm{f'}\left( x\right) =\frac{x}{\sqrt{{x}^{2}+1}}+\frac{3-x}{\sqrt{{x}^{2}-6\,x+18}}$
We need to find the values for $\mathrm{f'}\left( x\right) = 0$, so by summing and multiplying both sides I got to the equation:
$2x^4 - 12x^3 + 19x^2 -6x+18 = 0$
But I don't think the purpose should be to solve a 4th grade equation, there should be another way I'm missing..
|
Norbert and Ross Millikan have already suggested one slick solution, and Gerry Myerson the even slicker solution, but you can do it purely algebraically: you ran into trouble because your derivative isn’t quite right. It should be
$$f\,'(x)=\frac{x}{\sqrt{1+x^2}}-\frac{3-x}{\sqrt{18-6x+x^2}}\;,$$
with the second term negative from the chain rule applied to $3-x$. Setting that to $0$ and doing a little algebra, we get
$$\begin{align*}
x\sqrt{18-6x+x^2}&=(3-x)\sqrt{1+x^2}\;,\\
18x^2-6x^3+x^4&=(3-x)^2(1+x^2)\;,\\
18x^2-6x^3+x^4&=9-6x+10x^2-6x^3+x^4\;,\\
18x^2&=9-6x+10x^2\;,\\
8x^2+6x-9&=0\;,\text{ and}\\
(4x-3)(2x+3)&=0\;.
\end{align*}$$
The desired solution is clearly $x=\dfrac34$.
|
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|
Summation of $ \frac{1}{2} + \frac{3}{4} + \frac{7}{8} + \frac{15}{16} + \cdots$ till $n$ terms What is the pattern in the following?
*
*Sum to $n$ terms of the series: $$ \frac{1}{2} + \frac{3}{4} + \frac{7}{8} + \frac{15}{16} + \cdots$$
|
Here is the pattern:
\begin{align*}
\frac{1}{2} + \frac{3}{4} + \cdots &= \biggl(1-\frac{1}{2}\biggr) + \biggl(1-\frac{1}{2^2}\biggr) + \cdots \\\ &= (1+1+\cdots + 1) - \biggl(\frac{1}{2}+\frac{1}{2^2} + \cdots +\frac{1}{2^n}\biggr)
\end{align*}
|
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|
Polynomials in Fourier trigonometric series I'm successively integrating $x^{n} \cos{k x}$ for increasing values of positive integer n. I'm finding:
$\frac{\sin{kx}}{k}$,
$\frac{\cos{kx}}{k^2}+\frac{x\sin{kx}}{k}$,
$\frac{2 x \cos{kx}}{k^2}+\frac{\left(-2+k^2 x^2\right)sin{kx}}{k^3}$,
$\frac{3 \left(-2+k^2 x^2\right) \cos{kx}}{k^4}+\frac{x \left(-6+k^2 x^2\right) \sin(kx)}{k^3}$
Is there a name for the sequence of polynomials: $x$, $2x$, $k^2x^2-2$, $3(k^2x^2-2)$, $x(k^2x^2-6)$ ... ?
Here is more:
$\frac{\sin{kx}}{k}$
$\frac{\cos{kx}}{k^2}+\frac{x \sin{kx}}{k}$
$\frac{2 x \cos{kx}}{k^2}+\frac{\left(-2+k^2 x^2\right) \sin{kx}}{k^3}$
$\frac{3 \left(-2+k^2 x^2\right) \cos{kx}}{k^4}+\frac{x \left(-6+k^2 x^2\right) \sin{kx}}{k^3}$
$\frac{4 x \left(-6+k^2 x^2\right) \cos{kx}}{k^4}+\frac{\left(24-12 k^2 x^2+k^4 x^4\right) \sin{kx}}{k^5}$
$\frac{5 \left(24-12 k^2 x^2+k^4 x^4\right) \cos{kx}}{k^6}+\frac{x \left(120-20 k^2 x^2+k^4 x^4\right) \sin{kx}}{k^5}$
$\frac{6 x \left(120-20 k^2 x^2+k^4 x^4\right) \cos{kx}}{k^6}+\frac{\left(-720+360 k^2 x^2-30 k^4 x^4+k^6 x^6\right) \sin{kx}}{k^7}$
|
The polynomials are recursive in nature, and this behavior is most apparent when the given integral
$$
\int x^n \cos(k x)\,dx
$$
is viewed as the real part of the function
$$
F_{n,k}(x) = \int x^n e^{i k x}\,dx = \int x^n \cos(k x)\,dx + i\!\int x^n \sin(k x)\,dx.
$$
Using integration by parts twice we can derive an inhomogeneous recurrence relation for these functions,
$$
k^2 F_{n,k}(x) - i k (n-1) F_{n-1,k}(x) + (n-1) F_{n-2,k}(x) = e^{i k x}\left(x^{n-1}-i k x^n\right).
\tag{1}
$$
Here we can define the polynomials
$$
P_{n,k}(x) = k^{n+1} e^{-i k x} F_{n,k}(x),
$$
and then use $(1)$ to derive their recurrence relation
$$
P_{n,k}(x) - i (n-1) P_{n-1,k}(x) + (n-1) P_{n-2,k}(x) = (kx)^{n-1} - i (kx)^n.
\tag{2}
$$
Now, if we write
$$
k^{n+1} \int x^n \cos(k x)\,dx = A_{n,k}(x) \cos(kx) - B_{n,k}(x) \sin(kx),
$$
where $A_{n,k}(x)$ and $B_{n,k}(x)$ are polynomials, we have
$$
A_{n,k}(x) = \operatorname{Re} P_{n,k}(x) \qquad \text{and} \qquad B_{n,k}(x) = \operatorname{Im}\, P_{n,k}(x).
$$
The first few polynomials $P_{n,k}(x)$ are
$$
\begin{align}
P_{1,k}(x) &= 1-i k x \\
P_{2,k}(x) &= 2 i+2 k x-i k^2 x^2 \\
P_{3,k}(x) &= -6+6 i k x+3 k^2 x^2-i k^3 x^3 \\
P_{4,k}(x) &= -24 i-24 k x+12 i k^2 x^2+4 k^3 x^3-i k^4 x^4 \\
P_{5,k}(x) &= 120-120 i k x-60 k^2 x^2+20 i k^3 x^3+5 k^4 x^4-i k^5 x^5 \\
P_{6,k}(x) &= 720 i+720 k x-360 i k^2 x^2-120 k^3 x^3+30 i k^4 x^4+6 k^5 x^5-i k^6 x^6 \\
P_{7,k}(x) &= -5040+5040 i k x+2520 k^2 x^2-840 i k^3 x^3-210 k^4 x^4+42 i k^5 x^5+7 k^6 x^6-i k^7 x^7
\end{align}
$$
|
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|
What is wrong with my solution? $\int \cos^2 x \tan^3x dx$ I am trying to do this problem completely on my own but I can not get a proper answer for some reason
$$\begin{align}
\int \cos^2 x \tan^3x dx
&=\int \cos^2 x \frac{ \sin^3 x}{ \cos^3 x}dx\\
&=\int \frac{ \cos^2 x\sin^3 x}{ \cos^3 x}dx\\
&=\int \frac{ \sin^3 x}{ \cos x}dx\\
&=\int \frac{ \sin^2 x \sin x}{ \cos x}dx\\
&=\int \frac{ (1 -\cos^2 x) \sin x}{ \cos x}dx\\
&=\int \frac{ (\sin x -\cos^2 x \sin x) }{ \cos x}dx\\
&=\int \frac{ \sin x -\cos^2 x \sin x }{ \cos x}dx\\
&=\int \frac{ \sin x }{ \cos x}dx - \int \cos x \sin x dx\\
&=\int \tan x dx - \frac{1}{2}\int 2 \cos x \sin x dx\\
&=\ln|\sec x| - \frac{1}{2}\int \sin 2x dx\\
&=\ln|\sec x| + \frac{\cos 2x}{4} + C
\end{align}$$
This is the wrong answer, I have went through and back and it all seems correct to me.
|
$\int\tan x\,dx=\ln|\sec x|=-\ln|\cos x|+ k $since $-\ln |\cos x|+k= \ln|(\cos x)^-1|+k = \ln|\sec x| +k$
$$\frac{\cos2x}{4}= \frac{1}{2}\cos^2(x) - 1/4$$
$k-1/4 = \text{new constant } C $ and together you have the solution.
Your answer seems to be correct it is just manipulation or different approach to trig identities.
|
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|
Expanding logarithm into series How to expand $f(x)=\ln\left( x+\sqrt{1+x^2} \right)$ into series at $x_0=0$? I've tried using Taylor's formula but counting consecutive derivatives was inconvenient and I couldn't find the general formula.
|
This is a somewhat laborious expansion, but it has a nice closed formula.
We begin with
$$f(x) = \frac{1}{\sqrt{1+x^2}}$$
and note that you function $F$ is such that $F'(x) = f(x)$. We can make use of the Generalized Binomial Theorem, namely
$$(1+\mathrm x)^\alpha = \sum_{n=0}^\infty {\alpha \choose n}\mathrm x^n$$
In your case, set $\mathrm x= x^2$ and $\alpha = -\dfrac{1}{2}$.
$$(1+x^2)^{-1/2}=\frac{1}{\sqrt{1+x^2}}=\sum_{n=0}^\infty {-1/2 \choose n}x^{2n}$$
It is clear the most important calculation will be that of $$ c_n={-1/2 \choose n}$$
Writing this explicitly, it gives
$$\eqalign{
& {c_n} = \frac{{\left( { - \frac{1}{2}} \right)\left( { - \frac{1}{2} - 1} \right)\left( { - \frac{1}{2} - 2} \right) \cdots \left( { - \frac{1}{2} - n + 1} \right)}}{{n!}} \cr
& {c_n} = \frac{1}{{n!}}\prod\limits_{k = 0}^{n - 1} {\left( { - \frac{1}{2} - k} \right)} \cr
& {c_n} = \frac{1}{{n!}}{\left( { - 1} \right)^n}\prod\limits_{k = 0}^{n - 1} {\left( {\frac{1}{2} + k} \right)} \cr
& {c_n} = \frac{1}{{n!}}{\left( { - 1} \right)^n}\prod\limits_{k = 0}^{n - 1} {\left( {\frac{{2k + 1}}{2}} \right)} \cr
& {c_n} = \frac{1}{{n!}}{\left( { - \frac{1}{2}} \right)^n}\prod\limits_{k = 0}^{n - 1} {\left( {2k + 1} \right)} \cr} $$
Note that for $n=1$ the product is empty, thati is, it is $1$.
If we write the product explicitly, we get
$$\prod\limits_{k = 0}^{n - 1} {\left( {2k + 1} \right)} = 1 \cdot 3 \cdots \left( {2n - 3} \right)\left( {2n - 1} \right)$$
We can "complete" it by adjoining the even numers:
$$\prod\limits_{k = 0}^{n - 1} {\left( {2k + 1} \right)} = {2^n}n!\frac{{1 \cdot 3 \cdots \left( {2n - 3} \right)\left( {2n - 1} \right)}}{{{2^n}n!}} = \frac{{1 \cdot 2 \cdot 3 \cdot 4 \cdots \left( {2n - 3} \right)\left( {2n - 2} \right)\left( {2n - 1} \right)2n}}{{{2^n}n!}}$$
and get
$$\prod\limits_{k = 0}^{n - 1} {\left( {2k + 1} \right)} = \frac{{\left( {2n} \right)!}}{{{2^n}n!}} = \frac{1}{{{2^n}}}\frac{{\left( {2n} \right)!}}{{\left( {2n - n} \right)!}}$$
so that
$${c_n} = {\left( { - 1} \right)^n}\frac{1}{{{2^n}n!}}\prod\limits_{k = 0}^{n - 1} {\left( {2k + 1} \right)} = {\left( { - 1} \right)^n}\frac{1}{{{4^n}}}\frac{{\left( {2n} \right)!}}{{n!\left( {2n - n} \right)!}} = {\left( { - 1} \right)^n}\frac{1}{{{4^n}}}{2n \choose n}$$
Thus we have that
$$\frac{1}{{\sqrt {1 + {x^2}} }} = \sum\limits_{n = 0}^\infty {{c_n}} {x^{2n}} = \sum\limits_{n = 0}^\infty {{{\left( { - \frac{1}{4}} \right)}^n}}{2n \choose n} {x^{2n}}$$
Thus, from our previous considerations, we get
$$\log \left( {x + \sqrt {1 + {x^2}} } \right) = \int\limits_0^x {\frac{{dt}}{{\sqrt {1 + {t^2}} }}} = \sum\limits_{n = 0}^\infty {{{\left( { - \frac{1}{4}} \right)}^n}}{2n \choose n} \frac{{{x^{2n + 1}}}}{{2n + 1}}$$
|
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|
If $ \frac{1}{c} = \frac{1}{2a} + \frac{1}{2b} $, then either $ a \leqslant c \leqslant b$ or $ b \leqslant c \leqslant a $ For $a, b = 1, 2, 3, \cdots$, let $ \frac{1}{c} = \frac{1}{2a} + \frac{1}{2b} $. Then prove that either $ a \leqslant c \leqslant b$ or $ b \leqslant c \leqslant a $ holds.
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Hint $\ $ For $\rm\ A = 1/a,\ B = 1/b,\ C = 1/c\ $ it becomes obvious, viz.
$$\rm C = \frac{A+B}2\ \Rightarrow\ A \ge C \ge B\ \ or\ \ B \ge C \ge A$$
|
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|
Find the possible value from the following. Find the possible value from the following.
I'm not able to end up on a concrete note, as I'm unable to get the essence of question, still not clear to me.
$x$, $y$, $z$ are distinct reals such that $y=x(4-x)$, $z=y(4-y)$, $x=z(4-z)$. The possible values of $x+y+z$ is:
$$\begin{array}{l} A.\ 4 && C.\ 7 \\ B.\ 6 && D.\ 9 \end{array}$$
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Composing the functions, we get
$$
\begin{align}
0
&=x^8-16x^7+104x^6-352x^5+660x^4-672x^3+336x^2-63x\\
&=x(x-3)(x^3-7x^2+14x-7)(x^3-6x^2+9x-3)\tag{1}
\end{align}
$$
The roots $x=0$ and $x=3$ lead to indistinct $x$, $y$, and $z$.
$x^3-7x^2+14x-7$ has 3 real roots in $[0,4]$ whose sum is $7$ (the negative of the coefficient of $x^2$).
$x^3-6x^2+9x-3$ has 3 real roots in $[0,4]$ whose sum is $6$ (the negative of the coefficient of $x^2$).
$x$, $y$, and $z$ all satisfy $(1)$.
$t(4-t)$ rotates the roots of the cubics.
Thus, the possible values of $x+y+z$ are $6$ and $7$.
|
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|
Find the value of $(a^3 + b^3 + c^3)/(abc)$ if $a/b + b/c + c/a = 1$.
Find the value of $$\frac{a^3+b^3+c^3}{abc}\qquad\text{ if }\quad \frac ab + \frac bc + \frac ca = 1.$$
I tried using Cauchy's inequality but it was of no help. Please guide me.
$a, b, c$ are real.
|
There is not enough information to solve this problem. Clearing out denominators, your hypothesis is
$$a^2 c + a b^2 + b c^2 = abc \quad (1)$$
and your desired conclusion is
$$a^3+b^3+c^3=kabc \quad (2)$$
for some constant $k$.
Suppose, for the sake of contradiction, there were a $k$ such that $(1)$ implied $(2)$. Since the polynomial $a^3+b^3+c^3-abc$ is irreducible, this would mean that $a^2 c+a b^2+b c^2-abc$ would divide $a^3+b^3+c^3-kabc$. But the two polynomials are both cubics, so the only way for the first to divide the second is the first is a scalar multiple of the second, and it isn't.
It's also easy to generate points on $(1)$ and see that the ratio $(a^3+b^3+c^3)/(abc)$ isn't constant. Just choose random values for $a$ and $b$ and equation $(1)$ turns into a quadratic; solving that quadratic gives you some points to try. You'll see very quickly that nothing like this is true.
|
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|
How to find the product of n terms of an arithmetic progression where common difference is not unity? How can we find the product of $n$ terms of an arithmetic progression where common difference is not unity?
I just want to know the last $3$ digits of $7 \times 23 \times 39 \times \ldots \times 2071$ where common difference is $16$.
|
There are two (unrelated) questions here. An answer to the first question is
$$\prod_{k=1}^n(ak+b)=a^n\frac{\Gamma\left(n+1+\frac{b}a\right)}{\Gamma\left(1+\frac{b}a\right)}$$
To answer the second question, consider $n=\prod\limits_{k=0}^{129}(7+16k)$. Note that $7+16k=0\pmod{5}$ for every $k=3\pmod{5}$. Using this for $k=3$, $k=8$ and $k=13$ yields $n=0\pmod{125}$, that is, $n=125m$ for some integer $m$.
Likewise, $7+16k=-1\pmod{8}$ and the number of terms in the product which defines $n$ is $130$, which is even, hence $n=1\pmod{8}$. And $125=5\pmod{8}$ hence $n=5m\pmod{8}$, which implies $m=5n\pmod{8}$, that is, $m=5\pmod{8}$.
Finally, $n=125\cdot5\pmod{125\cdot8}$, that is, $n=\underline{\ \ \ \ \ \ \ \ }\pmod{1000}$.
|
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|
$\int_{0}^{\infty}\frac{\sin^{2n+1}(x)}{x} \mathrm {d}x$ Evaluate Integral Here is a fun integral I am trying to evaluate:
$$\int_{0}^{\infty}\frac{\sin^{2n+1}(x)}{x} \ dx=\frac{\pi \binom{2n}{n}}{2^{2n+1}}.$$
I thought about integrating by parts $2n$ times and then using the binomial theorem for $\sin(x)$, that is, using $\dfrac{e^{ix}-e^{-ix}}{2i}$ form in the binomial series.
But, I am having a rough time getting it set up correctly. Then, again, there is probably a better approach.
$$\frac{1}{(2n)!}\int_{0}^{\infty}\frac{1}{(2i)^{2n}}\sum_{k=0}^{n}(-1)^{2n+1-k}\binom{2n}{k}\frac{d^{2n}}{dx^{2n}}(e^{i(2k-2n-1)x})\frac{dx}{x^{1-2n}}$$
or something like that. I doubt if that is anywhere close, but is my initial idea of using the binomial series for sin valid or is there a better way?.
Thanks everyone.
|
I am just adding the proof of the identity for those who have interest:
$$ \sin^{2n+1} x = \frac{1}{4^n}\sum_{k=0}^{n}(-1)^{n-k}\binom{2n+1}{k}\sin\left(\left(2(n-k)+1\right)x\right). $$
Using the complex representation and the Binomial Theorem, we have
$$\begin{aligned}
\sin^{2n+1}x&=\left(\frac{\mathrm{e}^{ix}-\mathrm{e}^{-ix}}{2i}\right)^{2n+1}\\
&=\frac{(-1)^n}{2^{2n+1}i}\sum_{k=0}^{2n+1}\binom{2n+1}{k}\mathrm{e}^{i(2n+1-k)x}(-1)^k\mathrm{e}^{i(-kx)}\\
&=\frac{(-1)^n}{2^{2n+1}i}\sum_{k=0}^{2n+1}(-1)^k\binom{2n+1}{k}\mathrm{e}^{i(2(n-k)+1)x}\\
&=\frac{(-1)^n}{2^{2n+1}i}\sum_{k=0}^{2n+1}(-1)^k\binom{2n+1}{k}\left[\cos\left(\left(2(n-k)+1\right)x\right) + i\sin\left(\left(2(n-k)+1\right)x\right)\right]\\
&=\frac{(-1)^n}{2^{2n+1}}\sum_{k=0}^{2n+1}(-1)^k\binom{2n+1}{k}\left[\sin\left(\left(2(n-k)+1\right)x\right) - i\cos\left(\left(2(n-k)+1\right)x\right)\right]
\end{aligned}
$$
Now, observe that
$$\begin{aligned}
\sum_{k=0}^{2n+1} a_{k} &= \sum_{k=0}^{n}a_{k}+\sum_{k=n+1}^{n+n+1}a_{k}\\
&=\sum_{k=0}^{n}a_{k}+\sum_{k=0}^{n}a_{n+1+k}\\
&=\sum_{k=0}^{n}a_{k}+\sum_{k=0}^{n}a_{n+1+n-k}\\
&=\sum_{k=0}^{n}\left(a_{k}+a_{2n+1-k}\right)
\end{aligned}
$$
Apply with $a_{k}=(-1)^{k}\binom{2n+1}{k}\left[\sin\left(\left(2(n-k)+1\right)x\right) - i\cos\left(\left(2(n-k)+1\right)x\right)\right]$, so
$$\begin{aligned}
a_{2n+1-k}&=-(-1)^{k}\binom{2n+1}{2n+1-k}\left[-\sin\left(\left(2(n-k)+1\right)x\right) - i\cos\left(\left(2(n-k)+1\right)x\right)\right]\\
&=(-1)^{k}\binom{2n+1}{k}\left[\sin\left(\left(2(n-k)+1\right)x\right) + i\cos\left(\left(2(n-k)+1\right)x\right)\right].
\end{aligned}
$$
Then,
$$ a_{k}+a_{2n+1-k}=2(-1)^{k}\binom{2n+1}{k}\sin\left(\left(2(n-k)+1\right)x\right). $$
Therefore,
$$\begin{aligned} \sin^{2n+1} x&=\frac{1}{4^{n}}\sum_{k=0}^{n}(-1)^{n-k}\binom{2n+1}{k}\sin\left(\left(2(n-k)+1\right)x\right)\\
&=\frac{1}{4^n}\sum_{k=0}^{n}(-1)^k\binom{2n+1}{n-k}\sin\left((2k+1)x\right)\\
&=\frac{1}{4^n}\sum_{k=0}^{n}(-1)^k\binom{2n+1}{n+k+1}\sin\left((2k+1)x\right),
\end{aligned}$$
as desired.
|
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|
Proof: For all integers $x$ and $y$, if $x^3+x = y^3+y$ then $x = y$ I need help proving the following statement:
For all integers $x$ and $y$, if $x^3+x = y^3+y$ then $x = y$
The statement is true, I just need to know the thought process, or a lead in the right direction. I think I might have to use a contradiction, but I don't know where to begin.
Any help would be much appreciated.
|
We have
\begin{eqnarray*}
x^3+x=y^3+y&\Longleftrightarrow& (x^3-y^3)+(x-y)=0\\
&\Longleftrightarrow& (x-y)(x^2+y^2+xy+1)=0.
\end{eqnarray*}
Since $x^2+y^2+xy+1=(x+\frac{y}{2})^2+\frac{3}{4}y^2+1>0$, we get $x=y$.
The hypothesis $x,y$ are integer numbers is redundant.
|
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|
Showing that the last digit of $a$ and $a^{13}$ are the same
For $a \in \mathbb N$, show that the last digit of $a$ and $a^{13}$
are the same.
For example: $2^{13} = 8,192$
$7^{13} = 96,889,010,407$
|
The last digit of $a^2$ in the set $\{1,2,3,4,5,0\}$ is
$\{1, 4, 9, 6, 5, 0\}$ (due to redundancy, I omit $6\le a\le 9$). Then
for $a^4$ we get $\{1,6,1,6,5,0\}$, the same for $a^8,a^{12}$ and so forth...
Now multiply $a^4$ once more to get
$$
\{1,6,1,6,5,0\}\cdot\{1,2,3,4,5,10\}=
\{1\cdot 1\; ,6\cdot2\; ,3\cdot1\; ,4\cdot 6\; ,5\cdot 5\; ,0\cdot 10 \}_{\bmod 10}
=\{1,2,3,4,5,0\}
$$
(think of $\cdot$ as Hadamard product) so you already get your last digit back for $a^5$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/173201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 3
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|
Functional Equation $f\big(f(x)\big)=3x$ over natural nubers for strictly increasing $f$
If $f$ is a strictly increasing function from the naturals to the naturals, and $f\big(f(x)\big)=3x$, what are all values of $f(2012)$?
I have only proven that $f(3x)=3f(x)$ but that gets nowhere :(
|
Proposition. $f(1) = 2$, $f(2) = 3$.
Proof. $f(1)$ cannot be $1$ otherwise we would have
$$
3 = f(f(1)) = f(1) = 1
$$
So $f(1) > 1$ and since $f$ is strictly increasing, $f(x) > x$ for each $x$.
Being $3 = f(f(1)) > f(1)$, the only remaining possibility is $f(1) = 2$.
Finally, $f(2) = f(f(1)) = 3$. $\blacksquare$
Proposition. $f(3^n x) = 3^n f(x)$.
Proof. This is a direct consequence of the relation
$$
f(3x) = f(f(f(x))) = 3f(x)\; \blacksquare
$$
Proposition. If $3^n < x \leq 2\cdot 3^n$ then $f(x) = 3^n + x$.
Proof. It's sufficient to note that there are exactly $3^n$ numbers between $3^n$ and $2\cdot 3^n$, and exactly $3^n$ numbers between
$$
f(3^n) = 3^nf(1)= 2\cdot 3^n
$$
and
$$
f(2\cdot 3^n) = 3^nf(2) = 3^{n +1} \; \blacksquare
$$
Proposition. If $2\cdot 3^n < x \leq 3^{n + 1}$ then $f(x) = 3x - 3^{n + 1}$.
Proof. Since $3^n < x - 3^n \leq 2\cdot 3^n$, the previous proposition implies $f(x -3^n) = x$ and therefore
$$
f(x) = f(f(x - 3^n)) = 3x - 3^{n + 1} \; \blacksquare
$$
Being $2\cdot 3^6 < 2012 \leq 3^7$, from last proposition we can conclude
$$
f(2012) = 3\cdot 2012 - 3^7 = 3849
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/174268",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
Find the limit without L'Hôpital's theorem I'm trying to find $$ \lim_{ n\to\infty} { {n^2+1}\over {n^3+1}} \cdot {\frac {n} {1}}$$
I know the answer is $1$, but I can't remember how my professor found it so simply without using L'Hôpital's theorem.
Could you please show me the shortcut?
|
We consider $\lim_{ n\to\infty} { {n^2+1}\over n^3+1} \cdot {\frac n 1}$.
To do this, we note that $\lim_{ n\to\infty} { {n^2+1}\over n^3+1} \cdot {\frac n 1} = \lim \frac{n^3 + n}{n^3 + 1} = \lim \frac{n^3\left(1 + \frac{1}{n^2}\right)}{n^3\left( 1 + \frac{1}{n^3}\right)} = \lim \frac{1 + \frac{1}{n^2}}{1 + \frac{1}{n^3}} = 1$
Alternately, we could directly carry out long polynomial division. Divide $n^3 + n$ by $n^3 + 1$. We get $1 + \frac{n-1}{n^3 + 1}$, which clearly has limit $1$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/176733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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|
Probability density function of a quotient of two normal random variables I have the following expression:
$$R=\frac{\sigma_1^2\nu_1(t)-\sigma_2^2\nu_2(t)}{\sigma_1^2\nu_1(t)+\sigma_2^2\nu_2(t)}$$
where:
$$[\nu_1(t),\nu_2(t)]$$ are two independent normally distributed random variables.
My question is: how can I find an expression for the probability density function $(pdf)$ of $R$?
Thanks.
|
Let $\nu_1$ and $\nu_2$ be independent standard normal random variables. Then $U=\frac{\nu_1}{\nu_2}$ is well known to follow Cauchy distribution with pdf:
$$
f_U(u) = \frac{1}{\pi} \frac{1}{1+u^2}
$$
Let $X = \frac{\sigma_1^2 - U \cdot \sigma_2^2}{\sigma_1^2 + U \cdot \sigma_2^2}$. Assuming $\sigma_1>0$ and $\sigma_2 > 0$, it is evident that the mapping $u \mapsto \frac{\sigma_1^2 - u \cdot \sigma_2^2}{\sigma_1^2 + u \cdot \sigma_2^2}$ maps $\mathbb{R}\backslash \{ -\frac{\sigma_1^2}{\sigma_2^2} \}$ to $\mathbb{R}\backslash \{-1\}$. Indeed, for $x \not= -1$,
$$
\frac{\sigma_1^2 - u \cdot \sigma_2^2}{\sigma_1^2 + u \cdot \sigma_2^2} = x \qquad \implies \qquad u(x) = \frac{1}{1+x} \left( \frac{\sigma_1^2}{\sigma_2^2} - x \right)
$$
Thus we readily read off $f_X(x)$ from the measure:
$$ \begin{eqnarray}
\mathrm{d} F_U(u) &=& f_U(u) \mathrm{d} u = \frac{1}{\pi} \frac{|u^\prime(x)|}{1+u^2(x)} \mathrm{d}x = \frac{2}{\pi} \frac{\sigma_1^2 \sigma_2^2}{(1+x)^2 \sigma_2^4 + \sigma_1^4(1-x)^2} \mathrm{d}x \\
&=& \frac{2}{\pi} \frac{\sigma_1^2 \sigma_2^2}{\left(\sigma_1^4 + \sigma_2^4\right)\left(x - \frac{\sigma_1^4-\sigma_2^4}{\sigma_1^4+\sigma_2^4} \right)^2 + \frac{4 \sigma_1^4 \sigma_2^4}{\sigma_1^4+\sigma_2^4}} \mathrm{d}x = \mathrm{d}F_X(x)
\end{eqnarray}
$$
We therefore see that $X$ follows a Cauchy distribution with location parameters $\mu = \frac{\sigma_1^4-\sigma_2^4}{\sigma_1^4+\sigma_2^4}$ and scale parameter $\gamma = \frac{2 \sigma_1^2 \sigma_2^2}{\sigma_1^4 + \sigma_2^4}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/176850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Prove trigonometry identity for $\cos A+\cos B+\cos C$ I humbly ask for help in the following problem.
If
\begin{equation}
A+B+C=180
\end{equation}
Then prove
\begin{equation}
\cos A+\cos B+\cos C=1+4\sin(A/2)\sin(B/2)\sin(C/2)
\end{equation}
How would I begin the problem I mean I think $\cos C $ can be $\cos(180-A+B)$. But I am unsure what to do next.
|
$\cos A+\cos B+\cos C$
$=2\cos\frac{A+B}{2}\cos\frac{A-B}{2}+1-2\sin^2\frac{C}{2}$ as $\cos2x=1-2\sin^2x$
Now $\cos\frac{A+B}{2}=\cos\frac{180^\circ - C}{2}=\cos(90^\circ-\frac{C}{2})=\sin\frac{C}{2}$
So, $\cos A+\cos B+\cos C$ becomes $2\sin\frac{C}{2}\cos\frac{A-B}{2}+1-2\sin^2\frac{C}{2}$
$=1+2\sin\frac{C}{2}(\cos\frac{A-B}{2}-\sin\frac{C}{2})$
$=1+2\sin\frac{C}{2}(\cos\frac{A-B}{2}-\cos\frac{A+B}{2})$ replacing $\sin\frac{C}{2}$ with $\cos\frac{A+B}{2}$
$=1+2\sin\frac{C}{2}(2\sin\frac{A}{2}\sin\frac{B}{2})$ applying $\cos(x-y)-\cos(x+y)= 2 \sin x \sin y$
$=1+4\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/176892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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"answer_id": 2
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|
Finding the sum of series; sum of squares Is there any way to find the sum of the below series ?
$$\underbrace{10^2 + 14^2 + 18^2 +\cdots}_{41\text{ terms}}$$
I got asked this question in a competitive exam.
|
Let $$A=10^2+14^2+18^2+\dots$$ to 41 terms. Then $A=4B$, where $$B=5^2+7^2+9^2+\dots$$ to 41 terms. Let $$C=6^2+8^2+10^2+\dots$$ to 41 terms. Then $$B+C=5^2+6^2+7^2+\dots=(1^2+2^2+3^2+\dots)-(1^2+2^2+3^2+4^2)$$ where the sums go to $86^2$. Also, $C=4D$, where $$D=3^2+4^2+5^2+\dots=(1^2+2^2+3^2+\dots)-(1^2+2^2)$$ where the sums go to $43^2$. So you can calculate $D$ from Argon's last formula, then from $D$ you can get $C$; you can get $B+C$ from Argon, then $B$, then, finally, $A$.
|
{
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"url": "https://math.stackexchange.com/questions/179735",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Summation equation for $2^{x-1}$ Since everyone freaked out, I made the variables are the same.
$$
\sum_{x=1}^{n} 2^{x-1}
$$
I've been trying to find this for a while. I tried the usually geometric equation (Here) but I couldn't get it right (if you need me to post my work I will). Here's the outputs I need:
1, 3, 7, 15, 31, 63
If my math is correct.
|
In this instance, without explicitly using the formula for geometric series,
$$\begin{align*}
\sum_{x=1}^n 2^{x-1} &= 1 + 2 + 2^2 + 2^3 + \cdots + 2^{n-1}\\
&= 1 + (1 + 2 + 2^2 + 2^3 + \cdots + 2^{n-1}) - 1 &\text{add and subtract}~ 1\\
&= (1 + 1) + (2 + 2^2 + 2^3 + \cdots + 2^{n-1}) - 1 &\text{regroup}\\
&= 2 + (2 + 2^2 + 2^3 + \cdots + 2^{n-1}) - 1\\
&= (2 + 2) + (2^2 + \cdots + 2^{n-1}) - 1 &\text{regroup again}\\
&= 2^2 + (2^2 + 2^3 + \cdots + 2^{n-1}) - 1\\
&= (2^2 + 2^2) + (2^3 + \cdots + 2^{n-1}) - 1 &\text{regroup again}\\
&= 2^3 + (2^3 + \cdots + 2^{n-1}) - 1\\
&= \cdots &\text{lather, rinse, repeat}\\
&= 2^{n-1} + (2^{n-1}) - 1 &\text{nearly done}\\
&= 2^n - 1.
\end{align*}$$
Now that we know the form of the result, it is also possible
to prove the result
$$\sum_{x=1}^n 2^{x-1} = 2^n - 1$$
more formally by induction. Clearly,
the result holds when $n = 1$ since $2^0 = 2^1 - 1$.
Then, if the result holds for some positive integer $n$, we have
that
$$\sum_{x=1}^{n+1} 2^{x-1} = \sum_{x=1}^n 2^{x-1} + 2^n
= (2^n-1) + 2^n = 2^{n+1} - 1$$
and so the result holds for $n+1$ as well. Since we know
that the result holds when $n=1$, it follows by induction
that it holds for all positive integers $n$.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
How many integer solutions to a diophantine equation Starting with the equation:
$\frac{1}{a}+\frac{1}{b}=\frac{p}{10^n}$,
I reached the equation:
$10^{n-log(p)} = \frac{ab}{a+b}$.
Now given the positive integer $n$, for what integer values of $p$ would the value of:
$10^{n-log(p)}$,
be rational?
Also, given positive integers $n$ and $p$, how would we find positive integer solutions to $a$ and $b$ that satisfy the second equation, where:
$a ≤ b$,
And is it possible to determine, given $n$ and $p$, how many $a$ and $b$ solutions exist?
|
As @Qiaochu Yuan noted,
$10^{n-\log p}=\dfrac{10^n}{p}$ is always a rational number.
Suppose that a positive integer $n$ and an integr $p$ are given,
and let $a$ and $b$ to be integers
satisfying the above relation:
$$
\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{p}{10^n}
\Longleftrightarrow
\\
pab = 10^na + 10^nb
\Longleftrightarrow
\\
p^2ab = 10^npa + 10^npb
\Longleftrightarrow
\\
p^2ab - 10^npa - 10^npb = 0
\Longleftrightarrow
\\
p^2ab - 10^npa - 10^npb + 10^{2n} = 10^{2n}
\Longleftrightarrow
\\
(pa - 10^n) (pb - 10^n) = 10^{2n}
.
$$
notice that both of the factors
$(pa - 10^n)$ and $(pb - 10^n)$
are (positive or negative) divisor of $10^{2n}$,
so we can conclude that
there is a (positive or negative) divisor $d$ of $10^{2n}$;
such that:
$$
(pa - 10^n)=d
\ \ \ \ \ \
\text{and}
\ \ \ \ \ \
(pb - 10^n)=\dfrac{10^{2n}}{d}
\ \ \ \ \ \
\Longleftrightarrow
$$
$$
a = \dfrac {d+10^n} {p}
\ \ \ \ \ \
\text{and}
\ \ \ \ \ \
b = \dfrac { \dfrac{10^{2n}}{d} +10^n} {p}
\ \ \ \ \ \
.
$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "11",
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|
What is the number of real roots of the polynomial The number of distinct real roots of the equation
$x^9+x^7+x^5+x^3+x+1=0.$
|
Let $f(x)=x^9+x^7+x^5+x^3+x+1$, then $f^\prime (x)=9x^8+7x^6+5x^4+3x^2+1\geq1>0.$ Hence $f$ is strictly increasing, so has atmost one real root. Also $f(0)=1>0$ and $f(-1)=-4<0$ and so has a root between $-1$ and $0$.
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Proof of $(a^2b+b^2c+c^2a)(ab^2+bc^2+ca^2)>\frac{(ab+bc+ca)^3}{3}$ For positive real numbers $a$, $b$ and $c$, how do we prove that:
$$(a^2b+b^2c+c^2a)(ab^2+bc^2+ca^2)>\frac{(ab+bc+ca)^3}{3}$$
|
By Cauchy-Schwarz:
$(a^2b+b^2c+c^2a)(ab^2+bc^2+ca^2) \ge (a^{1.5}b^{1.5}+b^{1.5}c^{1.5}+c^{1.5}a^{1.5})^2$
By the power mean inequality:
$(a^{1.5}b^{1.5}+b^{1.5}c^{1.5}+c^{1.5}a^{1.5})^2 \ge \frac{(ab+bc+ca)^3}{3}.$
(Note: The > sign in your post is false; let $a=b=c.$)
|
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|
Finding all $x$ for $\frac{2x - 13}{2x + 3} \lt \frac{15}{x}$ I'm trying to find all $x$ for the inequality $\frac{2x - 13}{2x + 3} \lt \frac{15}{x}$. In order to do this, I want to factor one side so that I can find all values where $x$ determines the term to equal $0$.
$$\frac{2x - 13}{2x + 3} \lt \frac{15}{x} \iff \frac{x(2x - 13) - 15(2x+3)}{x(2x + 3)} \lt 0$$
$$\iff \frac{2x^2 - 13x - 30x - 45}{2x(x + \frac{3}{2})} \lt 0 \iff \frac{x^2 - \frac{43}{2}x - \frac{45}{2}x}{x(x + \frac{3}{2})} \lt 0 $$
$$\iff \frac{x^2 - \frac{43}{2}x - \frac{45}{2}x}{x(x + \frac{3}{2})} \lt 0 \iff \frac{(x - \frac{43}{4})^2 - \frac{1939}{4}}{x(x+\frac{3}{2})} \lt 0$$
I don't know how to get any further, and I'm starting to get too high values to handle.
The next step as I can see would be to find an $x$ that makes $(x - \frac{43}{4})^2 = \frac{1939}{4}$. This, alongside the obvious ones for $x$ and $x+\frac{3}{2}$ (creating division by $0$), would help me find the possible values for $x$.
But how do I get the last step? Or am I already dead wrong?
|
There is at least one incorrect "solution" above.
The usual trick is to multiply both sides by something positive. So multiply both sides by $x^2(2x+3)^2$. Next, eliminate common factors in the numerators and denominators, bring everything over to the left hand side and then factorise. The inequality reduces to
$x(2x-45)(2x+3)(x+1) < 0 \, .$
You need to find the values of $x$ for which an odd number of $x$, $2x-45$, $2x+3$ and $x+1$ are negative. So clearly $x = 0$, $x = 45/2$, $x = -3/2$ and $x = -1$ play an important role: these are the values for which the factors change sign. The final result, contrary to Iuli's "solution", is: $-3/2 < x < -1$ or $0 < x < 45/2$.
To see why Iuli's "solution" is incorrect, let $x = -1/2 \in (-3/2,45/2) \backslash \{0\}$, yet
$-7 \not< -30 \, . $
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Number of triples $(a,b,c)$ of positive integers such that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{3}{4}$?
What is the number of triples $(a,b,c)$ of positive integers such that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{3}{4}$ is:
A) $16$
B) $25$
C) $31$
D) $19$
E) $34$
Note: Through trial and error I've concluded that it's greater than $19$, but I wonder if there is a better way to solve it. I believe this question was on a contest, but I don't know which.
|
Assume that $a\le b\le c$. The average of the fractions $\frac1a,\frac1b$, and $\frac1c$ is $\frac14$, so $\frac1a\ge\frac14$, and $a\le 4$. Clearly $a>1$, so $a$ must be $2,3$, or $4$.
*
*If $a=4$, the only possibility is $a=b=c=4$.
*Suppose that $a=3$. If $b=3$, $\frac1c=\frac34-\frac23=\frac1{12}$, so $c=12$. If $b=4$, $\frac1c=\frac34-\frac13-\frac14=\frac16$, so $c=6$. If $b\ge 5$, then $c\ge 5$ as well, so $\frac1a+\frac1b+\frac1c\le\frac13+\frac25=\frac{11}{15}<\frac34$, so there are no more solutions with $a=3$.
*Suppose that $a=2$; clearly $\frac1b+\frac1c=\frac14$, so $b>4$. If $b=5$, $\frac1c=\frac14-\frac15=\frac1{20}$, so $c=20$. If $b=6$, $\frac1c=\frac14-\frac16=\frac1{12}$, so $c=12$. If $b=7$, then $\frac1c=\frac14-\frac17=\frac3{28}$, so there is no solution in this case. If $b=8$, clearly $c=8$ as well. Finally, if $b>8$, then $\frac1b+\frac1b\le\frac29<\frac14$, and there are no further solutions.
The solutions $\langle a,b,c\rangle$ with $a\le b\le c$ are therefore $\langle 4,4,4\rangle$, $\langle 3,3,12\rangle$, $\langle 3,4,6\rangle$, $\langle 2,5,20\rangle$, $\langle 2,6,12\rangle$, and $\langle 2,8,8\rangle$, for a total of six solutions. If you count different permutations of the same integers separately, there are $3\cdot3!+2\cdot3+1=25$ solutions.
|
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|
All pairs (x,y) that satisfy the equation $xy+(x^3+y^3)/3=2007$ How we can find the all pairs $(x,y)$ from the integers numbers ,that satisfy the equation :
$$xy+\frac{x^3+y^3}{3} =2007$$
|
Observe that the equation is symmetric.
As $3|(x^3+y^3)$, either $(x,y)$ will be $(3a+1,3b-1)$, $(3a-1,3b+1)$ or $(3a,3b)$.
If $(x,y)$ is $(3a+1,3b-1)$, $\frac{x^3+y^3}{3}=3(3a^3+3b^3+3a^2-3b^2+a+b)$
So, 3 must divide $xy$ which is impossible as $xy=(3a+1)(3b-1)$
So, $(x,y)$ will be $(3a,3b)$.
So,$9(ab+a^3+b^3)=2007\implies a^3+b^3+ab=223$
Now, 223 is prime, so, $(a,b)=1$
If we think of solution in natural number, $a<7$ .
By trial (which is aided by $(a,b)=1$), $(a,b)$ is $(6,1)$ or $(1,6)$.
|
{
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|
3 nonlinear equations in 3 variables - with the same variable raised to different powers Hoping someone can give me some guidance on solving the following system of 3 nonlinear equations in 3 variables:
$$\left\{\begin{align*}
&x^2+y^2=100\\
&xy+yz=-102\\
&y^2+z^2=117
\end{align*}\right.$$
Thanks!
|
Observe that $y ≠0$ for finite $x,z$
$z=0\implies y^2=117>100\implies x^2=-17\implies x^2z^2=117(-17)≠(-102)^2$
$x=0\implies y^2=100\implies z^2=17 \implies y^2z^2=1700≠(-102)^2$
So $xyz≠0$
$(3)-(1)\implies z^2-x^2=17\implies z+x=\frac{17}{z-x}$
$(2)\implies y(z+x)=-102\implies y=-\frac{102}{\frac{17}{z-x}}=6(x-z)$
Putting the value of $y$ in $(1),(3)$.
$x^2+36(x-z)^2=100\implies 37x^2-72zx+z^2=100-->(4)$
and $z^2+36(x-z)^2=117\implies 37z^2-72zx+x^2=117-->(5)$
On division of $(4)$ by $(5)$, $\frac{37x^2-72zx+z^2}{37z^2-72zx+x^2}=\frac{100}{117}$
$\implies x^2(37\cdot 117 -100)-72zx(117-100)+z^2(117-37\cdot 100)=0$
$\implies 4229x^2-1224zx-3583z^2=0$
If $x=a\cdot z, 4229a^2-1224a-3583=0-->(6)$ as $xyz≠0$
So, $y=6(x-z)=6(az-z)=6z(a-1)$
So, from (2), $az 6z(a-1)+6z(a-1)z=-102$
or $z^2=\frac{17}{1-a^2}$ where $a$ is a root of $(6)$.
|
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|
Showing that $ \frac{a}{a^2+3}+\frac{b}{b^2+3}\leq\frac{1}{2}$ for $a,b > 0$ and $ab = 1$ using rearrangement inequalities Please help to solve the following inequality using rearrangement inequalities.
Let $a \gt 0$, $b \gt0$ and $ab=1$. Prove that
\begin{equation}\frac{a}{a^2+3}+\frac{b}{b^2+3}\leq\frac{1}{2}.\end{equation}
Thanks.
|
An opportunity to show the defects of AM-GM, you get a weaker inequailty.
$\dfrac{a^2+3}{a}=a+\dfrac{3}{a} \ge 2 \sqrt{3} \implies (\dfrac{a^2+3}{a})^{-1} \le \dfrac{1}{2 \sqrt{3}}$
$\dfrac{b^2+3}{b}=b+\dfrac{3}{b} \ge 2 \sqrt{3} \implies (\dfrac{b^2+3}{b})^{-1} \le \dfrac{1}{2 \sqrt{3}}$
When you add them, you have weaker inequality. Therefore, re-arrangement inequality is an appropriate one!
|
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Automorphic numbers
Problem. We say that the $n$-digit number $x$ is automorphic iff $x^2\equiv x \mod(10^n)$. Prove that if $x$ is $n$-digit automorphic number then $(3x^2-2x^3)\mod(10^{2n})$ is $2n$-digit automorphic number. Hint: use Chinese reminder theorem to find the necessary and sufficient condition for number to be automorphic.
So from Chinese reminder theorem we have that $x$ is automorphic iff: $$ \begin{cases} x(x-1)\equiv 0 \mod(2^n)\\ x(x-1)\equiv 0 \mod(5^n)\end{cases} $$ which gives us four systems of equations:
$$ \begin{cases} x\equiv 0 \mod(2^n)\\ x\equiv 0 \mod(5^n)\end{cases} $$
$$ \begin{cases} x\equiv 1 \mod(2^n)\\ x\equiv 1 \mod(5^n)\end{cases} $$
$$ \begin{cases} x\equiv 0 \mod(2^n)\\ x\equiv 1 \mod(5^n)\end{cases} $$
$$ \begin{cases} x\equiv 1 \mod(2^n)\\ x\equiv 0 \mod(5^n)\end{cases} $$
and it's easy to check that thesis is true for the first two cases, just by simple operations. But how to check the last two?
|
To do the computation with modular arithmetic instead of divisibility....
If $x \equiv 0 \pmod{2^n}$, then $x \equiv x_1 2^n \pmod{2^{2n}}$ for some integer $x_1$. Then,
$$ 3x^2 - 2x^3 \equiv 3 x_1^2 2^{2n} - 2 x_1^3 2^{3n} \equiv 0 \pmod{2^{2n}}$$
Similarly, in the case of $x \equiv 1 + x_1 2^n \pmod{2^{2n}}$, then,
$$ \begin{align*} 3x^2 - 2x^3 &\equiv 3(1 + x_1 2^n)^2 - 2(1 + x_1 2^n)^3
\\& \equiv 3(1 + 2 x_1 2^n) - 2(1 + 3 x_1 2^n)
\\&\equiv 1 &\pmod{2^{2n}}\end{align*}$$
where this time I didn't bother writing the higher powers of $2$ that would appear but vanish modulo $2^{2n}$.
You should be made aware of the similarity to differential approximation: in fact, for any polynomial function $f(x)$, we have
$$ f(x_0 + x_1 p^n) \equiv f(x_0) + x_1 f'(x_0) p^n \pmod{p^{2n}} $$
(this can be generalized to Taylor series too. If this is interesting, look up "$p$-adic analysis")
If one had known this fact, one could have quickly solved the problem by setting $f(x) = 3 x^2 - 2 x^3$ and observing:
*
*$f(0) = 0$
*$f(1) = 1$
*$f'(0) = f'(1) = 0$
|
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|
Calculation of a strange series Is it possible to find an expression for:
$$S(N)=\sum_{k=0}^{+\infty}\frac{1}{\sum_{n=0}^{N}k^n}?$$
For $N=1$ we have
$$S(1) = \displaystyle\sum_{k=0}^{+\infty}\frac{1}{1 + k} = \displaystyle\sum_{k=1}^{+\infty}\frac{1}{k}$$
which is the (divergent) harmonic series. Thus, $S (1) = \infty$.
For $N=2$ this sum is:
$$S(2)=\sum_{k=0}^{+\infty}\frac{1}{1+k+k^2}$$
which can be expressed as:
$$S(2)=-1+\frac{1}{3}\sqrt 3 \pi \tanh(\frac{1}{2}\pi\sqrt 3)\approx 0.798$$
For $N=3$ we have:
$$S(3)=\frac{1}{4}\Psi(I)+\frac{1}{4I}\Psi(I)-\frac{1}{4I}\pi\coth(\pi)+\frac{1}{4}\pi\coth(\pi)+\frac{1/}{4}\Psi(1+I)-\frac{1}{4I}\Psi(1+I)-\frac{1}{2}+\frac{1}{2}\gamma \approx 0.374$$
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Perform a partial fraction decomposition:
$$
\frac{1}{p(k)} = \frac{1}{1+k+\cdots+k^{n-1}} = \frac{1}{ \prod_{m=1}^{n-1}\left(k-\exp\left(i \frac{2 \pi}{n} m \right)\right)} = \sum_{m=1}^{n-1} \frac{1}{k-\exp\left(i \frac{2 \pi}{n} m \right)} \frac{1}{p^\prime\left(\exp\left(i \frac{2 \pi}{n} m \right)\right)}
$$
Now:
$$
p^\prime\left(z\right) = \sum_{m=1}^{n-1} m z^{m-1} = \frac{\mathrm{d}}{\mathrm{d}z} \sum_{m=0}^{n-1} z^{m} = \frac{\mathrm{d}}{\mathrm{d}z} \frac{1-z^n}{1-z} = \frac{z-z^n (n-z(n-1))}{z (1-z)^2}
$$
Therefore, using $z^n=1$ for $z=\exp\left(i \frac{2 \pi}{n} m \right)$:
$$
c_m := \frac{1}{p^\prime\left(\exp\left(i \frac{2 \pi}{n} m \right)\right)} = \frac{1}{n} \exp\left(i \frac{2 \pi}{n} m \right) \left( \exp\left(i \frac{2 \pi}{n} m \right) - 1 \right) = \frac{1}{n} \exp\left(i \frac{2 \pi}{n} m \right) \left( \exp\left(i \frac{2 \pi}{n} m \right) - 1 \right)
$$
We thus have, and using $\sum_{m=1}^{n-1} c_m = 0$:
$$ \begin{eqnarray}
\sum_{k=0}^\infty \frac{1}{p(k)} &=& \sum_{k=0}^\infty \sum_{m=1}^{n-1} \frac{c_m}{k-\exp\left(i \frac{2 \pi}{n} m \right)} = \sum_{k=0}^\infty \sum_{m=1}^{n-1} c_m \left(\frac{1}{k-\exp\left(i \frac{2 \pi}{n} m \right)} - \frac{1}{k+1}\right) \\ &=& -\sum_{m=1}^{n-1} c_m \sum_{k=0}^\infty \left(\frac{1}{k+1} - \frac{1}{k-\exp\left(i \frac{2 \pi}{n} m \right)}\right) \\ &=& -\sum_{m=1}^{n-1} c_m \left( \gamma + \psi\left(-\exp\left(i \frac{2 \pi}{n} m \right)\right)\right)
\end{eqnarray}
$$
Again, making use of $\sum_{m=1}^{n-1} c_m = 0$ we arrive at:
$$
\sum_{k=0}^\infty \frac{1}{1+k+\cdots+k^{n-1}} = \sum_{m=1}^{n-1} \frac{1}{n} \exp\left(i \frac{2 \pi}{n} m \right) \left(1- \exp\left(i \frac{2 \pi}{n} m \right) \right) \cdot \psi\left(-\exp\left(i \frac{2 \pi}{n} m \right)\right)
$$
where $\psi(x)$ denotes the digamma function.
|
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Inequality. $a\sqrt{2b+c^2}+b\sqrt{2c+a^2}+c\sqrt{2a+b^2} \leq 3\sqrt{3}$ Could you help me please with the following inequality
Let $a$, $b$ and $c$ be non-negative numbers such that $a+b+c=3.$ Prove that:
$$a\sqrt{2b+c^2}+b\sqrt{2c+a^2}+c\sqrt{2a+b^2} \leq 3\sqrt{3}.$$
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Since $f(x)=\sqrt{x}$ is a concave function, by Jensen we obtain:
$$\sum_{cyc}a\sqrt{2b+c^2}=3\sum_{cyc}\frac{a}{3}\sqrt{2b+c^2}\leq3\sqrt{\sum_{cyc}\frac{a}{3}(2b+c^2)}=\sqrt{3\sum_{cyc}(2ab+a^2b)}=$$
$$=\sqrt{3(9-\sum_{cyc}(a^2-a^2b)}=\sqrt{27-\left((a+b+c)(a^2+b^2+c^2)-3\sum_{cyc}a^2b\right)}=$$
$$=\sqrt{27-\sum_{cyc}(a^3-2a^2b+a^2c)}=\sqrt{27-\sum_{cyc}(a^3-2a^2b+b^2a)}=\sqrt{27-\sum_{cyc}a(a-b)^2}\leq3\sqrt3$$
|
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How to evaluate $\frac{1}{b^2}\int_0^\infty z^{-2}\exp(-a z)\sin^2(b z)\, \mathrm dz$? How can I integrate the following:
$$\frac{1}{b^2}\int_0^\infty z^{-2}\exp(-a z)\sin^2(b z)\, \mathrm dz$$
for $a,b>0$? Maple gives a compact result:
$$\frac{1}{b} \tan^{-1}(c) - \frac{1}{ac^2} \ln(1 + c^2)$$
where $c=2b/a$. I can solve it when the powers of $z$ and $\sin$ are $-1$ and $1$ respectively, but not for 2.
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Let $I(a)$ denote the integral. Consider
$$\begin{eqnarray}
I^{\prime\prime}(a) &=& \frac{1}{b^2} \int_0^\infty \exp(-a z) \sin^2(b z) \mathrm{d}z = \frac{1}{2b^2} \int_0^\infty \exp(-a z) (1-\cos(2 b z)) \mathrm{d}z \\
&=& \frac{1}{2b^2} \Re \int_0^\infty (\exp(-a z)-\exp(-(a+ 2i b) z)) \mathrm{d}z = \frac{1}{2b^2} \Re\left(\frac{1}{a} - \frac{1}{a+2 i b}\right) = \frac{2}{a} \frac{1}{a^2 + 4 b^2}
\end{eqnarray}
$$
Now integrate back, using $I(\infty) = 0$ and $I^\prime(\infty)=0$:
$$
I^{\prime}(a) = \int_{a}^\infty \frac{2}{t} \frac{1}{t^2 + 4 b^2} \mathrm{d}t = \frac{1}{4 b^2} \log\left(1+ \frac{4b^2}{a^2}\right)
$$
$$
I(a) = \int_{a}^\infty I^{\prime}(t) \mathrm{d}t = \frac{1}{b} \arctan\left(\frac{2b}{a}\right) - \frac{a}{4b^2} \log\left(1+ \frac{4b^2}{a^2}\right)
$$
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|
How to prove this inequality $\sqrt{\frac{ab+bc+cd+da+ac+bd}{6}}\geq \sqrt[3]{{\frac{abc+bcd+cda+dab}{4}}}$ How to prove this inequality
$$\sqrt{\frac{ab+bc+cd+da+ac+bd}{6}}\geq \sqrt[3]{{\frac{abc+bcd+cda+dab}{4}}} $$
for $a,b,c,d\gt0$?
Thanks
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I remember seeing this problem in some book a couple of years ago. My attention was drawn by the proof of this inequality, which apparently originates from the '70 GDR mathematical olympiad. The solution involves nothing more than AM-GM (I assume $a,b,c,d$ are nonnegative here!), but some algebraic transformations are pretty insane, so stay calm $\ddot\smile$
\begin{equation}
\begin{split}
\quad &\sqrt{\dfrac{ab + ac + ad + bc + bd + cd}{6}} = \\
&\sqrt{\dfrac{(ab+cd)/2 + (ac+bd)/2 + (ad+bc)/2}{3}} \geq \quad \text{// AM-GM applied here} \\
&\sqrt[6]{\dfrac{(ab+cd)(ac+bd)(ad+bc)}{8}} = \\
&\sqrt[6]{\dfrac{a^3bcd + ab^3cd + abc^3d + abcd^3}{8} + \dfrac{a^2b^2c^2 + a^2b^2d^2 + a^2c^2d^2 + b^2c^2d^2}{8}} = \\
&\sqrt[6]{\dfrac{\left(\dfrac{a^2 + b^2}{2} + \dfrac{b^2 + c^2}{2} + \dfrac{c^2 + d^2}{2} + \dfrac{d^2 + a^2}{2}\right)abcd + \dfrac{a^2 + c^2}{2}b^2d^2 + \dfrac{b^2 + d^2}{2}a^2c^2}{8} + }
\\
&\hspace{120pt} \overline{+ \dfrac{a^2b^2c^2 + a^2b^2d^2 + a^2c^2d^2 + b^2c^2d^2}{16}} \geq \quad \text{// and here}\\
&\sqrt[6]{\dfrac{a^2b^2c^2 + a^2b^2d^2 + a^2c^2d^2 + b^2c^2d^2}{16} +} \\
&\hspace{120pt} \overline{+ \dfrac{2(a^2b^2cd + ab^2c^2d + abc^2d^2 + a^2bcd^2 + ab^2cd^2 + a^2bc^2d)}{16}} = \\
&\sqrt[6]{\left(\dfrac{abc + abd + acd + bcd}{4}\right)^2} = \sqrt[3]{\dfrac{abc + abd + acd + bcd}{4}} _{\square}
\end{split}
\end{equation}
Side note: if anyone knows how to improve the formatting (make the multi-line 6th root look smoother), feel free to edit this. :) I also used \overline to make the two long root expressions a bit more readable, I hope it renders properly.
|
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|
$(1+i)$ to the power $n$
Possible Duplicate:
Complex number: calculate $(1 + i)^n$.
I came across a difficult problem which I would like to ask you about:
Compute
$ (1+i)^n $ for $ n \in \mathbb{Z}$
My ideas so far were to write out what this expression gives for $n=1,2,\ldots,8$, but I see no pattern such that I can come up with a closed formula.
Then I remember that one can write any complex number $a+bi$ like:
$$(a+bi)=\sqrt{a^2+b^2} \cdot \left( \frac{a}{\sqrt{a^2+b^2}}+\frac{b}{\sqrt{a^2+b^2}}\cdot i\right)$$
and $\frac{a}{\sqrt{a^2+b^2}} = \cos(\phi)$ and $\frac{b}{\sqrt{a^2+b^2}} = \sin(\phi)$ where $\phi$ is $\arctan{\frac{b}{a}} $
So it becomes,
$(a+bi)=\sqrt{a^2+b^2} \cdot ( \cos(\phi)+\sin(\phi)\cdot i)$ Taking this entire thing to the power $n$ using De Moivre
$$(a+bi)^n=(\sqrt{a^2+b^2})^n \cdot ( \cos(n\phi)+\sin(n\phi)\cdot i)$$
Substituting my $a=1$ and $b=1$
$(1+i)^n=(\sqrt{2})^n \cdot ( \cos(n\cdot\frac{\pi}{4})+\sin(n\cdot\frac{\pi}{4})\cdot i)$
$\phi$ is 45 degrees hence $\frac{\pi}{4}$
But now I don't know how to continue further and I would really appreciate any help! Again, Im looking for a closed formula depending on n.
Best regards
|
I think you are making this much too difficult, because:
$$(1+i)^2 = 2i.$$
So $(1+i)^{2m} = (2i)^m = 2^mi^m$.
That takes care of the even powers of $1+i$.
For the odd powers, just multiply by $1+i$ again: $(1+i)^{2m+1} = (1+i)^{2m}(1+i) = 2^mi^m(i+1)$.
If you want to simplify further, you need to know the remainder when $m = \bigl\lfloor \frac n2 \bigr\rfloor$ is divided by 4, so that you can reduce the $i^m$ part.
Say $n$ is even, and $m=n/2 = 4k+j$, where $0\le j<3$; then $n=8k+2j$:
$$\begin{array}{ll}
j & (1+i)^{n=8k+2j} \\
\hline
0 & \hphantom{-}2^m \\
1 & \hphantom{-}2^mi \\
2 & -2^m \\
3 & -2^mi
\end{array}$$
If $n$ is odd, and $m=(n-1)/2 = 4k+j$, where $0\le j<3$, then $n=8k+2j+1$:
$$\begin{array}{ll}
j & (1+i)^{n=8k+2j+1} \\
\hline
0 & \hphantom{-}2^m+2^mi \\
1 & -2^m+2^mi \\
2 & -2^m-2^mi \\
3 & \hphantom{-}2^m-2^mi
\end{array}$$
Putting these together into one table, we can just say that $n=8k+j'$, where $0\le j'\lt 8$:
$$\begin{array}{lrl}
j' & (1+i)^{n=8k+j'} \\
\hline
0 & \hphantom{-}2^m & \\
1 & \hphantom{-}2^m&+2^mi \\
2 & &\hphantom{-}2^mi \\
3 & -2^m&+2^mi \\
4 & -2^m& \\
5 & -2^m&-2^mi \\
6 & &-2^mi \\
7 & \hphantom{-}2^m&-2^mi
\end{array}$$
(And remember $m=\bigl\lfloor \frac n2 \bigr\rfloor$.)
|
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|
Show $8\mid n^2-1$ if $n$ is an odd positive integer. Show that $n^2-1$ is divisible by $8$, if $n$ is an odd positive integer.
Please help me to prove whether this statement is true or false.
|
Since $n$ is odd $n=4m+1$ or $n=4m+3$.
In the first case $n^2-1=(n-1)(n+1)=4m\cdot(4m+2)=8m(2m+1)$, while in the second case $n^2-1=(n-1)(n+1)=(4m+2)\cdot(4m+4)=8(2m+1)(m+1)$.
So $n^2-1$ is divisible by 8 if $n$ is odd.
|
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|
A inequality proposed at Zhautykov Olympiad 2008 An inequality proposed at Zhautykov Olympiad 2008.
Let be $a,b,c >0$ with $abc=1$. Prove that:
$$\sum_{\mathrm{cyc}}{\frac{1}{(a+b)b}} \geq \frac{3}{2}.$$
Set $a=\frac{x}{y}$, $b=\frac{y}{z}$, $c=\frac{z}{x}$.
Our inequality becomes:
$$\sum_{\mathrm{cyc}}{\frac{z^2}{zx+y^2}} \geq \frac{3}{2}.$$
Now we use that: $z^2+x^2 \geq 2zx.$
$$\sum_{\mathrm{cyc}}{\frac{z^2}{zx+y^2}} \geq \sum_{\mathrm{cyc}}{\frac{2z^2}{z^2+x^2+2y^2}} \geq \frac{3}{2}.$$
Now applying Cauchy-Schwarz we obtain the desired result.
What I wrote can be found on this link: mateforum.
But now, I don't know how to apply Cauchy-Schwarz.
Thanks:)
|
Here is a very short one.
Suppose w.l.o.g. $a \ge b \ge c$. Then by the rearrangement inequality,
$$
S = \frac{1}{(a+b)b} +\frac{1}{(b+c)c} + \frac{1}{(a+c)a} \ge \frac{1}{(a+b)c} +\frac{1}{(b+c)a} + \frac{1}{(a+c)b} = T
$$
So
$$
2 S \ge S + T = \frac{b+c}{(a+b)bc} +\frac{c+a}{(b+c)ca} + \frac{a+b}{(a+c)ab} \geq 3 \Big( \frac{1}{abc} \Big)^{2/3} = 3
$$
which proves it, where in the last step AM-GM was used.
|
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|
A problem about multiples.
For any positive integers $a$, $ b$, if $ab+1$ is a multiple of $16$, then $a+b$ must be a multiple of $p$. Find the largest possible value of $p$.
I have no idea how to solve this. Please help. Thank you.
|
Let $ab\equiv-1\pmod {2^k}$
If $a=1,b\equiv-1= c2^k-1,a+b=c2^k$, the minimum value of $c$ is $1$.
So, $p$ must divide $2^k$, so can not have any odd factor.
Now, $b\equiv -\frac 1 a\pmod {2^k}$ as $ab\equiv-1$
So, $b\equiv -\frac 1 a\pmod {2^t}$
Let $2^t||(a+b)$ clearly, $t\le k$
So,$a\equiv -b \pmod{2^t}\equiv \frac 1 a \implies a^2\equiv 1\pmod{2^t}$ as $(a,2)=1$
As $a$ is odd, $=2d+1$ (say), $a^2-1=8\frac{d(d+1)}2\implies 8\mid (a^2-1) ∀$ odd $a$.
Now $\frac{d(d+1)}2$ will be even if $4\mid d(d+1)\implies d=4e$ or $d+1=4e$ as $(d,d+1)=1$
So, $\frac{d(d+1)}2$ will be odd if $d=4e+1$ or if $d=4e+2$
If $d=4e+1,a=2d+1=8e+3,a^2-1=64e^2+48e+8=8(8e^2+6e+1)$
If $d=4e+2,a=2d+1=8e+5,a^2-1=64e^2+80e+24=8(8e^2+10e+3)$
In a more compact way,if $a=8e\pm 3, a^2-1=(8e\pm 3)^2-1=64e^2\pm48c+8=8(8c^2\pm6c+1)$
$\implies 2^3||(a^2-1)$ if $a\equiv 3,5\pmod 8$
$\implies p=8 ∀ k\ge 3$ as $k=2\implies a+b=4c$
|
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|
Correctness of Fermat Factorization Proof I have asked similar questions regarding this proof. But now I would like to know if my reformulation (after perseverance and different thinking) is correct.
Prove: An odd integer $n \in \mathbb{N}$ is composite $\iff$ $n$ can be written as $n = x^2 - y^2 s.t. y+1 < x$
Proof: $\leftarrow$
Let $n$ be odd and consider $n= x^2-y^2 s.t. y+1 < x$
We can factor the difference of two squares as: $(x+y)(x-y)$ and we note for $n$ to be composite (not prime) $(x-y) > 1$. Thus we have shown this direction.
$\rightarrow$
Let $n$ be odd an composite. By definition of composite we have $n=ab$ for some odd integers $a$ and $b$. Now, working backwards:
$\dfrac{4ab}{4} = \dfrac{(a^2 + 2b + b^2)}{4} - \dfrac{(a^2-2b +b^2)}{4} = (\dfrac{a+b}{2})^2 - (\dfrac{a-b}{2})^2$. Thus, we shall take $x = \dfrac{a+b}{2}$ and $y =\dfrac{a-b}{2}$. And we have $n = x^2 - y^2$.
How do I get $y+1 < x$ in this direction?
|
Since $n$ is composite $a,b>1$ (WLOG $a\geq b$)
$$x-y=\dfrac{a+b}{2}-\dfrac{a-b}{2}=b>1$$
$$x>y+1$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluate $\sum_{k=1}^n\lfloor \sqrt{k} \rfloor$ I know that $\sum_{k=1}^m k^2 = \frac{m(m+1)(2m+1)}{6}$.
How can I use this fact to evaluate $\sum_{k=1}^n\lfloor \sqrt{k} \rfloor$ ?
|
When $i^2\leq k<(i+1)^2-1$, $\lfloor \sqrt{k} \rfloor=i$.
So $\lfloor \sqrt{k} \rfloor$ takes $2i+1$ times the value $i$.
If $m^2 \leq n < (m+1)^2$, $\sum_{k=1}^n \lfloor \sqrt{k} \rfloor= (\sum_{k=1}^{m^2-1} \lfloor \sqrt{k} \rfloor)+ (\sum_{k=m^2}^n \lfloor \sqrt{k} \rfloor)=
(\sum_{i=1}^{m-1} (2i+1) \times i) + (n-m^2+1) \times m=\sum_{i=1}^{m-1}2i^2+\sum_{i=1}^{m-1}i +(n-m^2+1)m$.
So the sum equals $m(m-1)(2m-1)/3+m(m-1)/2+(n-m^2+1)m$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/207320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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|
How to find the largest rectangle inside an ellipse I have an ellipse that is defined by center, width and height. The axes of the ellipse parallel to the x and y. I want to find the largest rectangle that completely fits inside this ellipse. Is there an easy way to do this?
And sorry if my terminology is a bit off... it's been a long time since I last tackled with geometry
|
Top right corner on the ellipse (centered on origin): $x = a \cos \theta$ and $y=b\sin\theta$
where $0\le \theta \le \frac{\pi}{2}, a=\frac{width}{2}$ and $b = \frac{height}{2}$
So area:$$A(\theta) = 4xy=4ab\sin\theta\cos\theta = 2ab\sin2\theta$$
Maximum of sin is at $2\theta=\frac{\pi}{2}$. So, we have $\theta = \frac{\pi}{4}$.
Therefore the 4 corners are:
$$(a\cos \frac{\pi}{4},b\sin \frac{\pi}{4}),
(a\cos \frac{3\pi}{4},b\sin \frac{3\pi}{4}),
(a\cos \frac{5\pi}{4},b\sin \frac{5\pi}{4}),
(a\cos \frac{7\pi}{4},b\sin \frac{7\pi}{4}),$$
which are of course equal to:
$$(\frac{a\sqrt{2}}{2},\frac{b\sqrt{2}}{2}),
(-\frac{a\sqrt{2}}{2},\frac{b\sqrt{2}}{2}),
(-\frac{a\sqrt{2}}{2},-\frac{b\sqrt{2}}{2}),
(\frac{a\sqrt{2}}{2},-\frac{b\sqrt{2}}{2}),$$
And it's area is: $a\sqrt{2}\cdot b\sqrt{2}=2ab$
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Elementary Question about limits Prove
$$
\lim_{x \to 0}\frac{x^3 - \sin^3x}{x - \ln{(1+x)} - 1 + \cos x} = 0
$$
Obviously, Using l'Hôpitals rule, we can evaluate this limit. But, taking derivatives of such functions is such a mess. Anyone sees a trick to do this problem faster? any ideas?
|
$$\dfrac{x^3 - \sin^3(x)}{x - \ln(1+x) - 1 + \cos(x)} = \dfrac{x - \sin(x)}{x} \times \dfrac{x^2 + x \sin(x) + \sin^2(x)}{1 - \dfrac{\ln(1+x)}{x}- \dfrac{1-\cos(x)}{x}}$$
$$1 - \dfrac{\ln(1+x)}{x}- \dfrac{1-\cos(x)}{x} = 1 - \left(1 - \dfrac{x}2 +\dfrac{x^2}3 - \cdots \right) - \left( \dfrac{\dfrac{x^2}{2!} - \dfrac{x^4}{4!} + \cdots}{x} \right) \\ = \left(\dfrac{x}2 - \dfrac{x^2}3 + \cdots \right) - \left(\dfrac{x}2 - \dfrac{x^3}{4!} + \cdots \right) = -\dfrac{x^2}3 + \mathcal{O}(x^4)$$
$$\dfrac{x - \sin(x)}{x} = \dfrac{x^2}{3!} + \mathcal{O}(x^4)$$
Hence,$$\dfrac{x^3 - \sin^3(x)}{x - \ln(1+x) - 1 + \cos(x)} = \left(\dfrac{x^2}{3!} + \mathcal{O}(x^4) \right) \times \dfrac{x^2 + x \sin(x) + \sin^2(x)}{-\dfrac{x^2}3 + \mathcal{O}(x^4)}$$
Now you should be able to finish it off.
|
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|
Invertible Matrix Let $A$ be the matrix
$$ A=\left(\begin{array}{cccc}1&0&1&2\\2&3&\beta&4\\4&0&-\beta&-8\\ \beta&0&\beta&\beta \end{array}\right). $$
For what values of $\beta$ is the matrix invertible?
|
$$\begin{vmatrix}
1&0&1&2\\
2&3&\beta&4\\
4&0&-\beta&-8\\
\beta&0&\beta&\beta
\end{vmatrix}
=
\beta\begin{vmatrix}
1&0&1&2\\
2&3&\beta&4\\
4&0&-\beta&-8\\
1&0&1&1
\end{vmatrix}
=
\beta\begin{vmatrix}
0&0&0&1\\
2&3&\beta&4\\
4&0&-\beta&-8\\
1&0&1&1
\end{vmatrix}
=
-\beta\begin{vmatrix}
2&3&\beta\\
4&0&-\beta\\
1&0&1
\end{vmatrix}
=3\beta
\begin{vmatrix}
4&-\beta\\
1&1
\end{vmatrix}
=3\beta(4+\beta)$$
So the determinant is zero only for $\beta=0$ and $\beta=-4$. (i.e.: these are the only cases when the matrix is singular.)
We have used the following:
*
*Effect of Elementary Row Operations on Determinant
*Laplace's Expansion Theorem
If - for some reason - you want to avoid determinants, you can simply do the elementary row operations.
(You mentioned in a comment that you haven't learned about determinants yet. It would have been better to mention this in your post.)
$$\begin{pmatrix}
1&0&1&2\\
2&3&\beta&4\\
4&0&-\beta&-8\\
\beta&0&\beta&\beta
\end{pmatrix}
\overset{(1)}\sim
\begin{pmatrix}
1&0&1&2\\
2&3&\beta&4\\
4&0&-\beta&-8\\
1&0&1&1
\end{pmatrix}
\sim
\begin{pmatrix}
0&0&0&1\\
2&3&\beta&4\\
4&0&-\beta&-8\\
1&0&1&1
\end{pmatrix}
\sim
\begin{pmatrix}
1&0&1&1\\
2&3&\beta&4\\
4&0&-\beta&-8\\
0&0&0&1
\end{pmatrix}
\sim
\begin{pmatrix}
1&0&1&1\\
0&3&\beta-2&2\\
0&0&-\beta-4&-12\\
0&0&0&1
\end{pmatrix}
\sim
\begin{pmatrix}
1&0&1&1\\
0&3&\beta-2&0\\
0&0&-\beta-4&0\\
0&0&0&1
\end{pmatrix}
\overset{(2)}\sim
\begin{pmatrix}
1&0&1&0\\
0&3&\beta-2&0\\
0&0&1&0\\
0&0&0&1
\end{pmatrix}
\sim
\begin{pmatrix}
1&0&0&0\\
0&3&0&0\\
0&0&1&0\\
0&0&0&1
\end{pmatrix}
\sim
\begin{pmatrix}
1&0&0&0\\
0&1&0&0\\
0&0&1&0\\
0&0&0&1
\end{pmatrix}
$$
You see that the matrix is invertible. But we have to be careful about steps $(1)$ and $(2)$. In $(1)$ we have divided by $\beta$ and in $(2)$ we have divided by $-\beta-4$. So these steps are valid only for $\beta\ne 0,4$. If you try elimination for $\beta=0$ and $\beta=4$, you will find out that for these values the matrix is singular.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Differentiating $x^2 \sqrt{2x+5}-6$ How do I differentiate this function: f(x)= $x^2 \sqrt{2x+5}-6$
I had: I had $2x\sqrt{2x+5} + x^2 \dfrac{1}{2\sqrt{2x+5}}$ but the correction model said it was I had $2x\sqrt{2x+5} + x^2 \dfrac{2}{2\sqrt{2x+5}}$
|
$2x . \sqrt{2x+5} + \dfrac{x^2}{2} . \dfrac{1}{\sqrt{2x+5}}2$
|
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|
For primes $p≡3\pmod 4$, prove that $[(p−1)/2]!≡±1\pmod p$. I know to use Wilson's Theorem and that each element in the second half is congruent to the negative of the first half, but I'm not sure how to construct a proof for it.
|
$p-r\equiv -r\pmod p\implies r\equiv-(p-r)$
For uniqueness, $r\le p-r$ or $2r\le p\implies r\le\frac p 2$
So, $1\le r\le \frac{p-1}2$ as $p$ is odd
Putting $r=1,2,3,\cdots,\frac{p-3}2,\frac{p-1}2$ we get,
$1\equiv-(p-1)$
$2\equiv-(p-2)$
...
$\frac{p-3}2\equiv-(p-\frac{p-3}2)=\frac{p+3}2$
$\frac{p-1}2\equiv-(p-\frac{p-1}2)=\frac{p+1}2$
So, there are $\frac{p-1}2$ pairs so,
$(p-1)!=(-1)^{\frac{p-1}2}\left((\frac{p-1}2)!\right)^2$
Using Wilson's theorem, $(-1)^{\frac{p-1}2}\left((\frac{p-1}2)!\right)^2\equiv-1\pmod p$
If $p\equiv3\pmod 4,p=4t+3$ for some integer $t$,
So, $\frac{p-1}2=2t+1$ which is odd, so $(-1)^{\frac{p-1}2}=-1$
$\implies \left((\frac{p-1}2)!\right)^2\equiv1\pmod p$
$\implies \left(\frac{p-1}2 \right)!\equiv\pm1\pmod p$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Inequality. $\frac{a^3}{a+b}+\frac{b^3}{b+c}+\frac{c^3}{c+a} \geq \frac{ab+bc+ca}{2}$ prove the following inequality:
$$\frac{a^3}{a+b}+\frac{b^3}{b+c}+\frac{c^3}{c+a} \geq \frac{ab+bc+ca}{2},$$ for $a,b,c$ real positive numbers.
Thanks :)
|
I started this proof but then realized that there was a mistake. Below is the wrong argument. Enjoy figuring out the mistake. (The argument can be fixed though.)
\begin{align}
S(a,b,c) & = \dfrac{a^3}{a+b} + \dfrac{b^3}{b+c} + \dfrac{c^3}{c+a}\\
& = \dfrac{a^3+b^3}{a+b} + \dfrac{b^3+c^3}{b+c} + \dfrac{c^3+a^3}{c+a} - \left(\dfrac{b^3}{a+b} + \dfrac{c^3}{b+c} + \dfrac{a^3}{c+a} \right)\\
& = (a^2 - ab + b^2) + (b^2 - bc + c^2) + (c^2 - ca + a^2) - S(a,c,b)
\end{align}
Hence, $$S(a,b,c) + S(a,c,b) = (a^2 - ab + b^2) + (b^2 - bc + c^2) + (c^2 - ca + a^2) \geq ab + bc+ ca$$
If $S(a,b,c) < \dfrac{ab + bc+ ca}2$, then $S(a,c,b) < \dfrac{ac + cb+ ba}2$ which gives us
$$S(a,b,c) + S(a,c,b) < ab + bc + ca$$ Contradiction.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
$(2^m -1)(2^n-1)$ divides $(2^{mn} -1)$ if and only if $\gcd(m,n) = 1$. If $\gcd(m,n) = 1$ then $(2^m-1)(2^n-1)$ divides $2^{mn} - 1$ because each of $2^m-1$, $2^n-1$ divide $2^{mn}-1$ and $\gcd(2^m-1, 2^n-1) = 2^{\gcd(m,n)}-1 = 1$. How about the converse? If $\gcd(m,n) > 1$, can one show that $(2^m-1)(2^n-1)$ does not divide $(2^{mn}-1)$?
|
The following generalization can be proved by a straightforward adaptation of the proof I gave above:
Given integer $a \geq 2$, positive integers $m,n$, and $k$ a positive integer that is a common divisor of $m$, $n$, then
$(a^m-1)(a^n-1)/(a^k-1)$ divides $a^{mn}-1$ if and only if $k = \gcd(m,n)$.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
sum of a series Can
\begin{equation}
\sum_{k\geq 0}\frac{\left( -1\right) ^{k}\left( 2k+1\right) }{\left(
2k+1\right) ^{2}+a^{2}},
\end{equation}
be summed explicitly, where $a$ is a constant real number? If $a=0,$ this
sum becomes
\begin{equation}
\sum_{k\geq 0}\frac{\left( -1\right) ^{k}}{2k+1}=\frac{\pi }{4}.
\end{equation}
What about for $a\neq 0$. I tried this method
\begin{eqnarray*}
\sum_{k\geq 0}\frac{\left( -1\right) ^{k}\left( 2k+1\right) }{\left(
2k+1\right) ^{2}+a^{2}} &=&\frac{1}{2}\sum_{k\geq 0}\left( -1\right) ^{k}%
\left[ \frac{1}{2k+1+ja}+\frac{1}{2k+1-ja}\right] \\
&=&\frac{1}{2}\sum_{k\geq 0}\frac{\left( -1\right) ^{k}}{2k+1+ja}+\frac{1}{2}%
\sum_{k\geq 0}\frac{\left( -1\right) ^{k}}{2k+1-ja} \\
&=&\frac{1}{2}\sum_{k\geq 0}\left( -1\right) ^{k}\int_{0}^{1}x^{2k+ja}dx+%
\frac{1}{2}\sum_{k\geq 0}\left( -1\right) ^{k}\int_{0}^{1}x^{2k-ja}dx \\
&=&\frac{1}{2}\int_{0}^{1}\left[ \sum_{k\geq 0}\left( -1\right) ^{k}x^{2k+ja}%
\right] dx+\frac{1}{2}\int_{0}^{1}\left[ \sum_{k\geq 0}\left( -1\right)
^{k}x^{2k-ja}\right] dx \\
&=&\frac{1}{2}\int_{0}^{1}\frac{x^{ja}}{1+x^{2}}dx+\frac{1}{2}\int_{0}^{1}%
\frac{x^{-ja}}{1+x^{2}}dx \\
&=&\frac{1}{2}\int_{0}^{1}\frac{x^{ja}+x^{-ja}}{1+x^{2}}dx
\end{eqnarray*}
but I was stucked at the last equations. Can any one give me some hint or tell me that the analytic expression doesn't exist. Thanks very much!
|
This sum can be evaluated by the same trick as presented here:
math.stackexchange.com.
In order to capture the convergence behavior of the series group consecutive terms to obtain absolute convergence, writing
$$ S(a) = \sum_{m\ge 0} \frac{4m+1}{(4m+1)^2+a^2}
- \sum_{m\ge 0} \frac{4m+3}{(4m+3)^2+a^2} =
\sum_{m\ge 0}
\frac{(4m+1)((4m+3)^2+a^2)-(4m+3)((4m+1)^2+a^2)}
{((4m+3)^2+a^2)((4m+1)^2+a^2)} =
\sum_{m\ge 0}
\frac{2(4m+1)(4m+3)-2a^2}{((4m+3)^2+a^2)((4m+1)^2+a^2)}
$$
Instead of using $f_1(z)$ and $f_2(z)$ from the other post use
$$f(z) =
\frac{2(4z+1)(4z+3)-2a^2}{((4z+3)^2+a^2)((4z+1)^2+a^2)} \pi \cot(\pi z).$$
The key operation of this technique is to compute the integral of $f(z)$ along a circle of radius $R$ in the complex plane, where $R$ goes to infinity. We certainly have $|\pi\cot(\pi z)|<2\pi$ for $R$ large enough. The core term from the sum is $\theta(R^2/R^4)$ which is $\theta(1/R^2)$ so that the integrals are $\theta(1/R)$ and vanish in the limit. This means that the sum of the residues at the poles of $f(z)$ add up to zero.
It is easily verified that the poles at the integers produce the terms of the sum twice over. It follows that
$$ 2 S(a)
+ \text{Res}_{z=\frac{3}{4}-\frac{1}{4}ia} f(z)
+ \text{Res}_{z=\frac{3}{4}+\frac{1}{4}ia} f(z)
+ \text{Res}_{z=\frac{1}{4}-\frac{1}{4}ia} f(z)
+ \text{Res}_{z=\frac{1}{4}+\frac{1}{4}ia} f(z)
= 0$$
The residues are easily computed as the poles are all simple.
This finally yields
$$ S(a) = \frac{1}{4} \frac{\pi}{\cosh \left(\frac{\pi a}{2}\right)}.$$
|
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|
The product of digits equal to the sum of digits How to find the number(or numbers ) that has $4$ digits, the product of these digits equal to the sum of these digits ?
|
First of all, let's observe that all of the digits of such a number cannot be the same. You can just manually check that numbers $1111$, $2222$ and so on don't suit us. It is also clear that all of the digits should be non-zero.
Now suppose that we have such a number. Let $a,\,b,\,c,\,d$ be its digits written in non-ascending order: $a \geqslant b \geqslant c \geqslant d$. Then we have
$$
abcd = a + b + c + d.
$$
From this we have an inequality:
$$
a\cdot bcd < 4a.
$$
This inequality is strict, because at least one of $b, c, d$ is strictly smaller than a. So we have:
$$
bcd < 4,
$$
which is the same as saying
$$
bcd \leqslant 3.
$$
This only leaves us with 3 possible combinations for $(b, c, d)$: $(1, 1, 1)$, $(2, 1, 1)$ and $(3, 1, 1)$.
If $b=c=d=1$, then $a\cdot 1 \cdot 1 \cdot 1 = a + 1 + 1 + 1$, which can't be true.
If $b=2$ and $c=d=1$, then $a \cdot 2 \cdot 1 \cdot 1 = a + 2 + 1 + 1$, which means that $a=4$. This gives us one possible solution: $a=4, b=2, c=d=1$.
If $b=3$ and $c=d=1$, then $a \cdot 3 \cdot 1 \cdot 1 = a + 3 + 1 + 1$, which is impossible.
So, the only solution is $a=4$, $b=2$, $c=d=1$. There are $12$ numbers with such digits.
|
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|
Middle line between points How can we calculate LINE that most fit the points T1(1,0) T2(2,0) T3(-1,1) T4(0,1)
$x= (1,2,-1,0) $
$y= (0, 0, 1, 1) $
$1= (1, 1, 1, 1)$
|
a)LINE that most fit the points $T1(1,0) \mspace{10mu} T2(2,0) \mspace{10mu} T3(-1,1) \mspace{10mu} T4(0,1) $
$x = (1 , 2 , -1 , 0) $
$y = (0 , 0 , 1 , 1) $
$1 = (1 , 1 , 1 , 1)$
Solving this: you got sistem of 2 equations:
$a < x, x> + b < 1, x> = < y, x> $
$a< x, 1> + b < 1, 1> = < y, 1>$
$< x, x>= x_1*x_1 + x_2*x_2 + x_3*x_3 + x_4*x_4 = 1*1 + 2*2 + (-1)*(-1) + 0*0 = 6 $
$< 1, x> = < x, 1>= 1*x_1 + 1*x_2 + 1*x_3 + 1*x_4 = 1*1 + 1*2 + 1*(-1) + 1*0 = 2$
$< x, 1>= x_1*1 + x_2*1 + x_3*1 + x_4*1 = 1*1 + 2*1 + (-1)*1 + 0*1 = 2 $
$< y, x>= x_1*y_1 + x_2*y_2 + x_3*y_3 + x_4*y_4 = 1*0 + 2*0 + (-1)*1 + 0*1= -1$
$< 1, 1> = 1 + 1 + 1 + 1 = 4 $
$< y, 1> = y_1*1 + y_2*1 + y_3*1 + y_4*1 = 0*1 + 0*1 + 1*1 + 1*1 = 2$
Then you insert these results in upper equation:
$6a + 2b = -1 $
$2a + 4b = 2$
$ a=\frac{-2}{5} and \mspace{10mu}
b=\frac{7}{10} $
Linear equation is:
$ \textbf{y = a x + b }$
Answer is:$ \mspace{10mu} \mathbf{y = \frac{-2}{5} + \frac{7}{10} }$
|
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|
Finding all real zeros of the polynomial $$x^5 - 5x^4 +6x^3 -30x^2 +8x - 40 = 0$$
So far I have...
$$r/s: +- 1, +- 40, +- 2, +- 20, +- 4, +- 10, +-5, +- 8$$
Only $+ 5$ works.
Then I have $$(x + 5)( ) = x^5 - 5x^4 +6x^3 -30x^2 +8x - 40$$
Then you have to use long devision between $x + 5$ and $ x^5 - 5x^4 +6x^3 -30x^2 +8x - 40$
That's where I get lost. Help?!
|
here is how to do the long division
x^5 - 5x^4 + 6x^3 - 30x^2 + 8x - 40
x^5 - 5x^4 subtract x^4(x-5)
-----------------------------------
6x^3 - 30x^2 + 8x - 40
6x^3 - 30x^2 subtract 6x^2(x-5)
-----------------------------------
8x - 40
8x - 40 subtract 8(x-5)
-----------------------------------
0
and since we got 0 remainder we can say that $$x^5 - 5x^4 + 6x^3 - 30x^2 + 8x - 40 = (x^4+6x^2+8)(x-5).$$
(Now you can find the roots of the quadratic $z^2+6z+8$ and square root them to find the real roots of the quartic)
|
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|
If $(a, b, c)$ is a Pythagorean triple, then $(a+1, b+1, c+1)$ is not a Pythagorean triple. Pythagoras stated that there exist positive natural numbers, $a$, $b$ and $c$ such that $a^2+b^2=c^2$. These three numbers, $a$, $b$ and $c$ are collectively known as a
Pythagorean triple. For example, $(8, 15, 17)$ is one of these triples as $8^2 + 15^2 = 64 + 225= 289 = 17^2$. Other examples of this triple are $(3, 4, 5)$ and $(5, 12, 13)$.
Using Proof by Contradiction, show that: If $(a, b, c)$ is a Pythagorean triple, then $(a+1, b+1, c+1)$ is not a Pythagorean triple.
|
Suppose both statement are true. Then
$(a+1)^2+(b+1)^2=(c+1)^2$
and as well
$a^2+b^2=c^2$
From 1.
$a^2+2a+1+b^2+2b+1=c^2+2c+1$
$\implies 2a+2b=2c-1$
But this does not satisfy for Pythagorean triplet 3,4,5 or for any other like 5,12,13.
Hence our supposition is wrong.
|
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|
Asymptotic behavior of $(1/2 + 2/3 + 3/4 + 4/5 + \cdots + (n-1)/n ) \times n$ I am interested in the following questions:
given:
$$G(n) = \left(\frac12 + \frac23 + \frac34 + \frac45 + \cdots + \frac{n-1}n\right)n$$
*
*what is a $F(n)$ which could be an upper bound (clearly as tight as possible) for $G(n)$ for $n$ arbitrarily large ?
*Does the series: $$\frac12 + \frac23 + \frac34 + \frac45 + \cdots + \frac{n-1}n$$ have a "name" and a sum (any reference)?
|
Let $$F(n) = \left(\dfrac{2-1}{2} + \dfrac{3-1}{3} + \dfrac{4-1}{4} + \cdots + \dfrac{n-1}{n} \right) $$ we then have that
$$F(n) = \left(1 - \dfrac12 + 1 - \dfrac13 + 1 - \dfrac14 + \cdots + 1 - \dfrac1n \right) = (n-1) - \left( \dfrac12 + \dfrac13 + \cdots + \dfrac1n \right)$$
Now note that $$-\left( \dfrac12 + \dfrac13 + \cdots + \dfrac1n \right) = 1 - H_n = 1 - \left(\log(n) + \gamma + \dfrac1{2n} - \dfrac1{12n^3} + \mathcal{O}(1/n^5) \right)$$
Hence, $$F(n) = n - \left(\log(n) + \gamma + \dfrac1{2n} - \dfrac1{12n^3} + \mathcal{O}(1/n^5) \right)$$
$$nF(n) = n^2 - n \log n -\gamma n - \dfrac12 + \dfrac1{12n^2} + \mathcal{O}(1/n^4)$$
You could make use of the fact that $$H_n = \log (n) + \gamma - \dfrac{\zeta(0)}n + \sum_{k=1}^{\infty} \dfrac{\zeta(-k)}{n^{k+1}}$$ to get better approximations/bounds.
|
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|
Integration by parts, Apostol Integration by parts. Need to show that:
If $I_n(x)=\int_{0} ^{x}t^n(t^2+a^2)^{-\frac{1}{2}}dt$
Then: $nI_n(x) = x^{n-1}\sqrt{x^2+a^2}-(n-1)a^2I_{n-2(x)}$ if $x\geq2$
I can get to the point where:
$I_n=\frac{x^{n+1}}{n+1}(x^2+a^2)^{-\frac{1}{2}}+\frac{1}{n+1}\int_0 ^x (t^nt^2+a^2t^n-a^2t^n)(t^2+a^2)^{-\frac{3}{2}}dt$
$=~\frac{x^{n+1}}{n+1}(x^2+a^2)^{-\frac{1}{2}}+ \frac{1}{n+1}I_n-\frac{1}{n+1}\int_0 ^x a^2t^n(t^2+a^2)^{-\frac{3}{2}}dt$
$=>~nI_{n}=x^{n+1}(x^2+a^2)^{-\frac{1}{2}}-a^2I_{n-2}+a^4\int t^{n-2}(t^2+a^2)^{-\frac{3}{2}}dt$
Now, is this a good start or should I have taken another route? Because I can not find a way out :/
|
Let $$I_n(x) = \int_0^x \dfrac{t^n}{\sqrt{t^2 + a^2}} dt = \int_0^x \dfrac{t^{n-1} \cdot t}{\sqrt{t^2 + a^2}} dt$$
Let $u(t) = t^{n-1}$ and $dv(t) = \dfrac{t}{\sqrt{t^2 + a^2}} dt \implies v(t) = \sqrt{t^2+a^2}$. Hence,
\begin{align}
I_n(x) & = \int_0^x u(t) dv(t) = \left. u(t) v(t) \right\vert_{0}^x - \int_0^x v(t)du(t)\\
& = x^{n-1} \sqrt{x^2 + a^2} - (n-1)\int_0^x t^{n-2} \sqrt{t^2+a^2} dt
\end{align}
\begin{align}
\int_0^x t^{n-2} \sqrt{t^2+a^2} dt & = \int_0^x \dfrac{t^{n-2} (t^2+a^2)}{\sqrt{t^2+a^2}} dt = \int_0^x \dfrac{t^{n}}{\sqrt{t^2+a^2}} dt + a^2 \int_0^x \dfrac{t^{n-2}}{\sqrt{t^2+a^2}} dt\\
& = I_n(x) + a^2 I_{n-2}(x)
\end{align}
Plugging this in, we get that
$$I_n(x) = x^{n-1} \sqrt{x^2 + a^2} - (n-1)(I_n(x) + a^2 I_{n-2}(x))$$
Rearranging, we get that
$$nI_n(x) = x^{n-1} \sqrt{x^2 + a^2} - (n-1)a^2 I_{n-2}(x)$$
|
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|
Solution for the value of angle $A$ of a triangle In triangle $\triangle \; ABC$ , if $$2\frac{\cos A}{a} + \frac{\cos B}{b} + 2\frac{\cos C}{c} = \frac{a}{bc} + \frac{b}{ca}$$ find angle $A$.
This is a quiz bee problem sent to me by my friend in FB. He asked me if I can do a solution for it. Well I tried several ways but I am out of idea now. The answer is 90 degree but what he asked, and I am also asking it now, is the solution for it.
Thank you.
|
Using $$\cos C=\frac{a^2+b^2-c^2}{2ab}$$ etc.,
we get, $$\frac2a\frac{b^2+c^2-a^2}{2bc}+\frac1b\frac{a^2+c^2-b^2}{2ac}+\frac2c\frac{a^2+b^2-c^2}{2ab}=\frac{a^2+b^2}{abc}$$
or, $$ 2(b^2+c^2-a^2)+(a^2+c^2-b^2)+2(a^2+b^2-c^2)=2(a^2+b^2)$$
or $b^2+c^2=a^2$ as $abc\ne0$ $a,b,c$ being the sides of triangle.
So, $\cos A=0\implies A=(2n+1)\frac\pi2 $ where $n$ is any integer.
As $0<A<\pi,A=\frac\pi2$
|
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|
$2n+1$ and $n^2+1$ are always coprime or their gcd is $5$ Using a spreadsheet, it can be inferred that when $n≡2[5]$, then $\gcd(n^2+1,2n+1)=5$, else $\gcd(n^2+1,2n+1)=1$.
Indeed, when $n≡2[5]$, $n^2+1$ and $2n+1$ can easily be shown to be multiples of $5$, so their gcd is at least $5$. But then, I can't see how to complete the proof.
|
As $(a,b)=(a,a-nb),$
$(2(n^2+1), 2n+1)=(2(n^2+1)-n(2n+1), 2n+1)=(2-n,2n+1)=(2-n,2n+1+2(2n-1))=(2-n,5)$
So, $(2(n^2+1), 2n+1)=5$ if $5\mid(2-n)$ i.e., if $n\equiv2\pmod 5$
$\implies (n^2+1, 2n+1)=5$ as $(2,2n+1)=1$ is as $(2n+1)$ odd.
If $n\not\equiv2\pmod 5,(2(n^2+1), 2n+1)=1\implies (n^2+1, 2n+1)=1$
Alternatively,
We know, $(a,b)\mid (ax+by)$ where $a,b,x,y$ are integers.
So, $(n^2+1, 2n+1)\mid \{n(2n+1)-2(n^2+1)\}\implies (n^2+1, 2n+1)\mid(n-2)$
Again, $(2n+1,n-2)\mid\{2n+1-2(n-2)\} \implies (2n+1,n-2)\mid 5$
If $n\equiv2\pmod 5,2n+1\equiv0 \pmod 5$ and $n^2+1\equiv2^2+1\equiv 0 \pmod 5$
hence, $(n^2+1, 2n+1)=5$
Else, $5\not\mid(n-2)\implies (2n+1,n-2)=1$ and $5\not\mid(2n+1)\implies (n^2+1, 2n+1)=1$
|
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|
Parametric Equations: Finding Given Tangent Lines At Points The parametric equations are: $x=2\cot\theta$ and $y=2\sin^2\theta$
$$\frac{dy}{dx}=-2\sin^3\theta\cdot \cos\theta$$
And the coordinates are: $(-\frac{2}{\sqrt{3}},3/2)$, $(0,2)$, and $(2\sqrt{3}, 1/2)$
I wasn't quite sure what to do with those coordinates, could someone please make sense of what I am to do, and why it should be done that way?
|
For example, with the point $\,\displaystyle{\left(-\frac{2}{\sqrt 3}\,,\,\frac{3}{2}\right)}\,$ , and with $\,t=\theta\,$ , for simplicity:
$$2\cot t=x=-\frac{2}{\sqrt 3}\Longrightarrow \tan t=-\sqrt 3\Longrightarrow t=\frac{\pi}{3}+k\pi\,\,,\,k\in\Bbb Z$$
$$\frac{3}{2}=y=2\sin^2t\Longrightarrow \sin t=\pm\frac{\sqrt 3}{2}\Longrightarrow t=\frac{\pi}{3}+k\pi\,\,\,,\,\,k\in\Bbb Z$$
Well, choose one of the infinite ammount of possible $\,t'$s above, say
$$t=\frac{\pi}{3}\,\,(k=0\,)\Longrightarrow\,\left.\frac{dy}{dx}\right|_{t=\pi/3}=-2\sin^3\frac{\pi}{3}\cos\frac{\pi}{3}=-2\left(\frac{\sqrt 3}{2}\right)^3\frac{1}{2}=-\frac{3\sqrt 3}{8}$$
so the tangent line to the curve at this point is
$$y-\frac{3}{2}=-\frac{3\sqrt 3}{8}\left(x+\frac{2}{\sqrt 3}\right)$$
|
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|
What am I doing wrong in calculating this determinant? I have matrix:
$$
A = \begin{bmatrix}
1 & 2 & 3 & 4 \\
2 & 3 & 3 & 3 \\
0 & 1 & 2 & 3 \\
0 & 0 & 1 & 2
\end{bmatrix}
$$
And I want to calculate $\det{A}$, so I have written:
$$
\begin{array}{|cccc|ccc}
1 & 2 & 3 & 4 & 1 & 2 & 3 \\
2 & 3 & 3 & 3 & 2 & 3 & 3 \\
0 & 1 & 2 & 3 & 0 & 1 & 2 \\
0 & 0 & 1 & 2 & 0 & 0 & 1
\end{array}
$$
From this I get that:
$$
\det{A} = (1 \cdot 3 \cdot 2 \cdot 2 + 2 \cdot 3 \cdot 3 \cdot 0 + 3 \cdot 3 \cdot 0 \cdot 0 + 4 \cdot 2 \cdot 1 \cdot 1) - (3 \cdot 3 \cdot 0 \cdot 2 + 2 \cdot 2 \cdot 3 \cdot 1 + 1 \cdot 3 \cdot 2 \cdot 0 + 4 \cdot 3 \cdot 1 \cdot 0) = (12 + 0 + 0 + 8) - (0 + 12 + 0 + 0) = 8
$$
But WolframAlpha is saying that it is equal 0. So my question is where am I wrong?
|
The method that you're using works just fine for $3\times 3$ matrices, but fails to work with $n\times n$ matrices for other $n$. You're going to have to do it another way.
For example, expanding the deteriminant along the first column, we find that $$\begin{align}\det A &=1\cdot\det\left[\begin{array}{ccc}3 & 3 & 3\\1 & 2 & 3\\0 & 1 & 2\end{array}\right]-2\cdot\det\left[\begin{array}{ccc}2 & 3 & 4\\1 & 2 & 3\\0 & 1 & 2\end{array}\right]+0\cdot\det\left[\begin{array}{ccc}3 & 3 & 3\\2 & 3 & 4\\0 & 1 & 2\end{array}\right]-0\cdot\det\left[\begin{array}{ccc}3 & 3 & 3\\2 & 3 & 4\\1 & 2 & 3\end{array}\right]\\ &= \det\left[\begin{array}{ccc}3 & 3 & 3\\1 & 2 & 3\\0 & 1 & 2\end{array}\right]-2\det\left[\begin{array}{ccc}2 & 3 & 4\\1 & 2 & 3\\0 & 1 & 2\end{array}\right].\end{align}$$
At that point, you can use your method of calculating determinants of $3\times 3$ matrices to get the rest of the way.
|
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|
Limit $\lim_{n \rightarrow \infty}\frac{1 \cdot 3 \cdot 5 \dots \cdot (2n-1)}{2 \cdot 4 \cdot 6 \cdot \dots \cdot (2n)} $ $\displaystyle \lim_{n \rightarrow \infty}\frac{1 \cdot 3 \cdot 5 \dots \cdot (2n-1)}{2 \cdot 4 \cdot 6 \cdot \dots \cdot (2n)} $
|
A hint:
Write $1\cdot3\cdot 5\cdot\ldots\cdot(2n-1)$ and $2\cdot 4\cdot 6\cdot\ldots\cdot (2n)$ in terms of factorials and powers of $2$. Then use Stirling's formula
$$m!=\left({m\over e}\right)^m\ \sqrt{2\pi m}\ \bigl(1+o(1)\bigr)\qquad(m\to\infty)$$
to estimate the various factorials appearing in your expression.
|
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|
A five-digit number N.... A five-digit number $N$ is equal to $45$ times the product of its $5$ digits. Find $N$.
Please help. I am not sure how to solve this. I have a feeling it is simple
|
Let the digits of $n$ from left to right be $a,b,c,d$, and $e$, so that we’re told that
$$10^4a+10^3b+10^2c+10d+e=45abcde\;.\tag{1}$$
Clearly $n$ is a multiple of $5$, so $e$ is either $0$ or $5$. But obviously none of the digits is $0$, so $e=5$, and $(1)$ becomes
$$10^4a+10^3b+10^2c+10d+5=225abcd$$
or, after dividing through by $5$,
$$2\left(10^3a+10^2b+10c+d\right)+1=45abcd\;.\tag{2}$$
The lefthand side of $(2)$ is odd, so all of $a,b,c$, and $d$ must be odd as well. The one’s digit of $$2\left(10^3a+10^2b+10c+d\right)+1$$ is either $2d+1$ or $2d+1-10=2d-9$, depending on whether $d<5$ or not. However, the number in $(2)$ is an odd multiple of $5$, so its one’s digit is $5$, and either $2d+1=5$ or $2d-9=5$. Since $d$ is known to be odd, the former is impossible, $d=7$, and $(2)$ becomes
$$20\left(10^2a+10b+c\right)+15=315abc$$
or, after another division by $5$,
$$4\left(10^2a+10b+c\right)+3=63abc\;.\tag{3}$$
Reduce $(3)$ modulo $9$ to see that $$4\left(10^2a+10b+c\right)\equiv 6\pmod 9$$ and hence $10^2a+10b+c\equiv6\pmod9$, which further implies that $a+b+c\equiv6\pmod9$. We know that $a,b$, and $c$ are odd, so $a+b+c$ is odd, and therefore $a+b+c$ can only be $15$; the next larger possibility is $33$, which is too large, since $a+b+c\le3\cdot9=27$. The largest of $a,b$, and $c$ must therefore be at least $5$.
Now note that $100,000>n=45abcde=1575abc$, so $$abc\le\left\lfloor\frac{100,000}{1575}\right\rfloor=63\;.$$ This reduces the possibilities for $a,b$, and $c$ to $7,7,1$ (in some order) and $9,5,1$ (in some order). In the first case we have $45abcde=77175$, so $77175$ is a solution. In the second case we have $45abcde=70875$ and get no solution. The unique solution is therefore $n=77175$.
|
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If $x$, $y$, $x+y$, and $x-y$ are prime numbers, what is their sum?
Suppose that $x$, $y$, $x−y$, and $x+y$ are all positive prime numbers. What is the sum of the four numbers?
Well, I just guessed some values and I got the answer.
$x=5$, $y=2$, $x-y=3$, $x+y=7$. All the numbers are prime and the answer is $17$.
Suppose if the numbers were very big, I wouldn't have got the answer.
Do you know any ways to find the answer?
|
Note that $x>y$, since $x-y$ is positive. Since $x$ and $y$ are both prime, this means that $x$ must be greater than $2$ and therefore odd. If $y$ were odd, $x+y$ would be an even number greater than $2$ and hence not prime. Thus, $y$ must be even, i.e., $y=2$.
Now we want an odd prime $x$ such that $x-2$ and $x+2$ are both prime. In other words, we want three consecutive odd numbers that are all prime. But one of $x-2,x$, and $x+2$ is divisible by $3$, so in order to be a prime it must be $3$. Clearly that one must be $x-2$, the smallest of the three numbers, and we have our unique solution: $x=5$ and $y=2$, and $x+y+(x+y)+(x-y)=3x+y=17$.
|
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How can I find the points at which two circles intersect? Given the radius and $x,y$ coordinates of the center point of two circles how can I calculate their points of intersection if they have any?
|
Easy solution is to consider another plane such that the centers are along an axis.
Given the points $(x_1,y_1)$ and $(x_2,y_2)$.
We focus on the center point of both circles given by
$$
\left( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} \right).
$$
The distance between the centers of the circles is given by
$$
R = \sqrt{ (x_2-x_1)^2 + (y_2-y_1)^2}.
$$
We can consider the following orthogonal vectors
$$
\vec{a} = \left( \frac{x_2-x_1}{R}, \frac{y_2-y_1}{R} \right),
\vec{b} = \left( \frac{y_2-y_1}{R}, - \frac{x_2-x_1}{R} \right).
$$
In the $(\vec{a},\vec{b})$ plane we get the equations
$$
\big( a + R / 2 \big)^2 + b^2 = r_1^2,\\
\big( a - R / 2 \big)^2 + b^2 = r_2^2.
$$
Whence
$$
a = \frac{r_1^2 - r_2^2}{2R},\\
b = \pm \sqrt{ \frac{r_1^2+r_2^2}{2} - \frac{(r_1^2-r_2^2)^2}{4R^2}
- \frac{R^2}{4}}.
$$
The intersection points are given by
$$
(x,y) =
\frac{1}{2} \big( x_1+x_2, y_1+y_2 \big)
+ \frac{r_1^2 - r_2^2}{2R^2} \big( x_2-x_1, y_2-y_1 \big)\\
\pm \frac{1}{2} \sqrt{ 2 \frac{r_1^2+r_2^2}{R^2} - \frac{(r_1^2-r_2^2)^2}{R^4}
- 1} \big( y_2-y_1, x_1-x_2 \big),
$$
where $R$ is the distance between the centers of the circles.
|
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|
Rotation matrix in terms of axis of rotation How to calculate the rotation matrix in 3D in terms of an arbitrary axis of rotation? Given a unit vector $V=V_{x}e_{x}+V_{y}e_{y}+V_{z}e_{z}$ How to calculate the rotation matrix about that axis?
|
I think you need the Rodrigue's rotation matrix composition.
If your unit rotation axis is $\vec{v} = (V_x,V_y,V_z)$ and the rotation angle $\theta$ then the rotation matrix is
$$ R = \boldsymbol{1}_{3\times3} + \vec{v}\times\,(\sin\theta) + \vec{v}\times\vec{v}\times\,(1-\cos\theta) $$
where $\vec{v}\times = \begin{pmatrix} 0 & -V_z & V_y \\ V_z & 0 & -V_x \\ -V_y & V_x & 0 \end{pmatrix}$ is the $3\times 3$ cross product operator matrix.
For example a rotation about the unit $\vec{v}=(\frac{\sqrt{3}}{3},0,\text{-}\frac{\sqrt{6}}{3})$ the rotation matrix is
$$ R = \begin{pmatrix} 1&0&0\\0&1&0\\0&0&1\end{pmatrix} +\begin{pmatrix} 0&\frac{\sqrt{6}}{3}&0\\ \text{-}\frac{\sqrt{6}}{3}&0&\frac{\sqrt{3}}{3}\\0&\frac{\sqrt{3}}{3}&0\end{pmatrix} \sin\theta + \begin{pmatrix} \text{-}\frac{2}{3}&0&\frac{\sqrt{2}}{3}\\0&\text{-}1&0\\\text{-}\frac{\sqrt{2}}{3}&0&\text{-}\frac{1}{3}\end{pmatrix} (1-\cos\theta) $$
which collects to:
$$ R = \frac{1}{3} \begin{pmatrix} 1\cos\theta+1 & \sqrt{6}\sin\theta & \sqrt{2}\cos\theta-\sqrt{2} \\ -\sqrt{6}\sin\theta& 3 \cos\theta & -\sqrt{3}\sin\theta \\ \sqrt{2}\cos\theta - \sqrt{2} & \sqrt{3}\sin\theta& \cos\theta+2 \end{pmatrix} $$
|
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|
How to get from $\frac{x}{x+1}\;$ to $\;1 - \frac{1}{x+1}$? Please show me how to manipulate $\dfrac{x}{x+1}\;\;$ to get $\;\;1 - \dfrac{1}{x+1}$
|
How about start with $1 - \frac{1}{x+1}$ to get
$$1 - \frac{1}{x+1} = \frac{x+1}{x+1} - \frac{1}{x+1} = \frac{x}{x+1}$$
|
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|
Maximization of an integer input function Maximize the value of the function
$$
z=\frac{ab+c}{a+b+c},
$$
where $a,b,c$ are natural numbers and are all lesser than 2010 and not necessarily
distinct from each other.
Please provide a proof, and if possible a general technique.
Thank you.
|
$$z=\frac{ab+c}{a+b+c}=\frac{ab-a-b}{a+b+c}+1$$
Clearly, this will be maximum if $c$ minimum $=1$
So, $$z\le \frac{ab-a-b}{a+b+1}+1=\frac{ab+1}{a+b+1} $$
$\frac{a_1b_1+1}{a_1+b_1+1}$ will be greater than $\frac{a_2b_2+1}{a_2+b_2+1}$
if $(a_1a_2-1)(b_1-b_2)+(b_1b_2-1)(a_1-a_2)+a_1b_1-a_2b_2>0$
if $(a_1a_2-1)(b_1-b_2)+(b_1b_2-1)(a_1-a_2)+(a_1-a_2)b_1+a_2(b_1-b_2)>0$
if $(a_1a_2-1+a_2)(b_1-b_2)+(b_1b_2-1+b_1)(a_1-a_2)>0$
Clearly, $\frac{ab+1}{a+b+1} $ increases with the increment of $a,b$ or both.
So, $\frac{ab+1}{a+b+1} $ will be maximum if $a,b$ are maximum.
|
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|
Expectation of joint life span The life span of a particular mechanical part is a random variable described by the following PDF:
If three such parts are put into service independently at t=0, determine a simle expression for the expected value of the time until the majority of the parts will have failed.
I can get the PDF:
$$
f_L(l) = 0.4 (0 \leq l \leq 2) \\
f_L(l) = -0.4l + 1.2 (2 < l \leq 3)
$$
and the expectation:
$$
E(l) = \int_0^3 l f_L(l) dl \approx 1.27
$$
I think 'majority' means 2 or more, so we can focus on two parts of the three, and pay no attention to the third. The translation is $E(max(l1, l2))$, how will this be derived I currently have no idea.
Sorry about the misleading remark "$E(max(l_1, l_2))$", it's wrong to neglect the third part, because if that one fails early, then we only need one of the rest to fail.
|
The value for $E \left[ \max \left( L_1, L_2 \right) \right]$ is computed in
the following way. First, the distribution of the maximum of two identically
independently distribued random variable $L_1$ and $L_2$ is given by $2 f
\left( \ell \right) F \left( \ell \right)$ where $f \left( \ell \right)$ is
the density and $F \left( \ell \right)$ is the cumulative distribution
function. This is well known, you could find the formula here. It is not
difficult to derive:
\begin{eqnarray*}
\Pr \left[ \max \left( L_1, L_2 \right) \leqslant \ell \right] & = & \Pr
\left[ \left\{ L_1 \leqslant \ell \right\} \cap \left\{ L_2 \leqslant \ell
\right\} \right]\\
& = & \Pr \left[ L_1 \leqslant \ell \right] \Pr \left[ L_2 \leqslant \ell
\right]\\
& = & F \left( \ell \right)^2
\end{eqnarray*}
Taking derivative gives the density $2 f \left( \ell \right) F \left( \ell
\right)$.
The probability density function $f(\ell)$ is given by (as you indicated)
$$ f \left( \ell \right) = \frac{2}{5} 1_{\ell} \left[ 0, 2 \right) + \left(
- \frac{2}{5} \ell + \frac{6}{5} \right) 1_{\ell} \left[ 2, 3 \right) $$
where the notation $1_{\ell}A$ with interval $A$ is that of an indicator variable. This means
$$ 1_{\ell} \left( A \right) = \left\{ \begin{array}{lll}
1 & & \text{if } \ell \in A\\
0 & & \text{otherwise}
\end{array} \right. $$
Therefore the cumlative distribution function is given by
$$ F( \ell)= \frac{2 \ell}{5} 1_{\ell} \left[ 0, 2
\right) + \frac{1}{5} \left( - \ell^2 + 6 \ell - 4 \right) 1_{\ell} \left[
2, 3 \right) + 1_{\ell} \left[ 3, \infty \right) $$
Multiplying both we get the density
\begin{eqnarray*}
2 f \left( \ell \right) F \left( \ell \right) & = & 2 \left\{ \frac{2}{5}
1_{\ell} \left[ 0, 2 \right) + \left( - \frac{2}{5} \ell + \frac{6}{5}
\right) 1_{\ell} \left[ 2, 3 \right) \right\}\\
& \times & \left\{ \frac{2 \ell}{5} 1_{\ell} \left[ 0, 2 \right) +
\frac{1}{5} \left( - \ell^2 + 6 \ell - 4 \right) 1_{\ell} \left[ 2, 3
\right) + 1_{\ell} \left[ 3, \infty \right) \right\}\\
& = & \frac{8 \ell}{25} 1_{\ell} \left[ 0, 2 \right) + \frac{4}{25} \left(
- \ell + 3 \right) \left( - \ell^2 + 6 \ell - 4 \right) 1_{\ell} \left[ 2, 3
\right)
\end{eqnarray*}
and therefore
\begin{eqnarray*}
E \left[ \max \left( L_1, L_2 \right) \right] & = & \frac{8}{25} \int_0^2
\ell^2 \mathrm{d} \ell + \frac{4}{25} \int_2^3 \ell \left( - \ell + 3
\right) \left( - \ell^2 + 6 \ell - 4 \right) \mathrm{d} \ell\\
& = & \frac{637}{375}\\
& \approx & 1.69867
\end{eqnarray*}
|
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|
Solutions of Diophantine equations in Natural numbers The one of solution of $x^4 - 2y^2 = -1$ is $x = 1$ and $y = 1$. However, the solution $(1, 1)$ of $x^4 - 2y^2 = 1$ is failed. We know $x = 1$ and $y = 1$ is small integers and we can check by trail method. In case more solutions are existing or not how to check? What are the solutions of $x^4 - 2y^2 = 1$?
|
The only integer solutions of $x^4-2y^2=1$ are $x=\pm 1$, $y=0$.
For $x$ must be odd. Rewrite our equation as $(x^2-1)(x^2+1)=2y^2$. The greatest common divisor of $x^2-1$ and $x^2+1$ is $2$. Since $x^2+1$ has shape $8k+2$, it follows that $x^2+1=2s^2$, and $x^2-1=t^2$ for some integers $s$ and $t$.
The only solution in integers of $x^2-1=t^2$ is $x=\pm 1$, $t=0$, since $1$ and $0$ are the only perfect squares that differ by $1$.
|
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|
Solutions for $ 2^i = 3^ j - 1 $ or $ 2^i = 3^ j + 1 $ Are there any solutions for $ 2^i = 3^ j - 1 $ or $ 2^i = 3^ j + 1 $,
for $i>3$ and $j>2$ ?
Thanks! $:)$
|
Observe that $2$ is a primitive root of $3^k$ where $k$ is natural number(Proof below).
(1)So if $2^i=3^{j+1}+1,2^i\equiv1\pmod {3^{j+1}}$
$\implies \phi(3^{j+1})\mid i \implies 2\cdot 3^j\mid i$
So, the minimum possible integral value of $i$ is $2\cdot 3^j$
If $j=0,i_{min}=2,2^i-3^{j+1}=2^2-3^1=1$
If $j=1,i_{min}=2\cdot3^1=6,2^i-3^{j+1}=2^6-3^2>1$
If $j=2,i_{min}=2\cdot3^2=18,2^i-3^{j+1}=2^{18}-3^3>1$
Clearly, the difference diverges from $1$ with the increment of $j$ and $i$.
It can also proved as follows:
$2^i-3^{j+1}-1=2^{2\cdot 3^j}-3^{j+1}-1=4^{3^j}-3^{j+1}$
$=(3+1)^{3^j}-3^{j+1}-1=\sum_{1\le k\le 3^j-1}\binom{3^j}k3^j>0$ if $j>0$
(2) So if $2^i=3^{j+1}-1,2^i\equiv-1\pmod {3^{j+1}}$
$\implies \frac{\phi(3^{j+1})}2\mid i$ but $\phi(3^{j+1})\not\mid i$
So, the minimum possible integral value of $i$ is $3^j$
If $j=0,i_{min}=1,2^i-3^{j+1}=2^1-3^1=-1$
If $j=1,i_{min}=3,2^i-3^{j+1}=2^3-3^2=-1$
If $j=2,i_{min}=3^2=9,2^i-3^{j+1}=2^9-3^{2+1}>-1$
Clearly, the difference diverges from $-1$ with the increment of $j$ and $i$.
[
Proof:
$2\equiv -1\pmod 3, 2^2\equiv1\implies ord_32=2$
$2^2=4,2^3=8\equiv-1\pmod {3^2},2^6\equiv1\implies ord_{(3^2)}2=6=2\cdot 3=\phi(9)$
From this, $ord_{(3^3)}2=2\cdot 3^2=\phi(27)$
As $ord_{(3^2)}2=6=2\cdot 3,ord_{(3^3)}2=2\cdot 3^2\implies ord_{(3^4)}2=2\cdot3^3=\phi(81)$
Proceeding this way, we can prove $ord_{(3^n)}2=2\cdot3^{n-1}=\phi(3^n)$
]
|
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|
Evaluating the infinite product $\prod\limits_{k=2}^\infty \left ( 1-\frac1{k^2}\right)$
Evaluate the infinite product
$$\lim_{ n\rightarrow\infty }\prod_{k=2}^{n}\left ( 1-\frac{1}{k^2} \right ).$$
I can't see anything in this limit , so help me please.
|
Hint: The "typical" term is $\dfrac{k-1}{k}\dfrac{k+1}{k}$. Express the first few terms in this way, and observe the nice cancellations.
For example, here is the product of the first seven terms:
$$\frac{1}{2}\frac{3}{2}\frac{2}{3}\frac{4}{3}\frac{3}{4}\frac{5}{4}\frac{4}{5}\frac{6}{5}\frac{5}{6}\frac{7}{6}\frac{6}{7}\frac{8}{7}\frac{7}{8}\frac{9}{8}=\frac{9}{2\cdot 8}.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
find recurrence relation $T(n)=2T(n/2) +\log_2(n)$ $$\begin{align*}
&T(n) = 2T(n/2) + \log_2(n)\\
&T(1) = 0
\end{align*}$$
$n$ is a power of $2$
solve the recurrence relation
my work so far:
unrolling this, we have
$$\begin{align*}
T(n) &= 4T(n/4) + \log_2(n) -1\\
&= 8T(n/8) + 2\log_2(n) -2\\
&=\log_2(n-1) \log_2(n) - \log_2(n) + 1
\end{align*}$$
after substituting for base case.
where is my mistake?
|
Substituting $n=2^k$ we have:
$$\begin{align*}T(n)=T(2^k)&=2T(2^{k-1})+k=2(2T(2^{k-2})+k-1)+k=4T(2^{k-2})+3k-1\\
&=4(2T(2^{k-3})+k-3)+3k-1=8T(2^{k-3})+7k-13=...=\\
&= 2^mT(2^{k-m})+\sum_{t=1}^m2^{t-1}(k-t+1)=...=2^kT(1)+\sum_{t=0}^{k-1}2^t(k-t)\\
&=2^k\cdot0+k\sum_{t=0}^{k-1}2^t-\sum_{t=0}^{k-1}t2^t=k\frac{2^k-1}{2-1}-2\frac{(k-1)2^k-k2^{k-1}+1}{2-1}\\
&=k2^k-k-2(k-1)2^k+k2^k-2=2k2^k-2k2^k+2\cdot2^k-k-2=\\
=&2\cdot2^k-k-2=2n-\log_2n-2\end{align*}$$
|
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|
trace of the matrix $I + M + M^2$ is
Let $ \alpha = e^{\frac{2\pi \iota}{5}}$ and the matrix
$$ M= \begin{pmatrix}1 & \alpha & \alpha^2 & \alpha^3 & \alpha^4\\
0 & \alpha & \alpha^2 & \alpha^3 & \alpha^4\\
0 & 0 & \alpha^2 & \alpha^3 & \alpha^4 \\
0 & 0 & 0 & \alpha^3 & \alpha^4\\
0 & 0 & 0 & 0 & \alpha^4 \end{pmatrix}$$
Then the trace of the matrix $I + M + M^2$ is
*
*$-5$;
*$0$;
*$3$;
*$5$.
I am stuck on this problem. Can anyone help me please?
I got trace of the matrix $$\operatorname{tr}(I+M+M^2) = 7 + \alpha + 2 \alpha^2 + \alpha^3 + 2 \alpha^4 + \alpha^6 +\alpha^8.$$ Now what to do?
|
Note that the trace of $M$ is $0$, since $1+\alpha+\alpha^2+\alpha^3+\alpha^4= 0$.
Also $M$ is upper triangular so that $M^2$ has diagonal elements which are just the square of the diagonal elements of $M$, i.e. $1,\alpha^2, \alpha^4, \alpha^6, \alpha^8$.
Using the fact that $\alpha^5 = 1$ we see that the trace of $M^2$ is again $0$.
Thus tr$(I+M+M^2)$ = tr$(I)$ = 5.
|
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|
Why do we choose $3$ to be positive after $\sqrt{9 - x^2}$ in the following substitution? The integral $$\int \frac{\sqrt{9 - x^2}}{x^2}dx$$ is solved in my book by letting $x = 3\sin\theta$ where $-\frac {\pi}{2} \le \theta \le \frac {\pi}{2}$.
Then, $dx = 3\cos\theta\,d\theta$
and, $$\sqrt{9-x^2} = 3|\cos\theta| = 3\cos\theta$$
So, $$\int \frac{\sqrt{9 - x^2}}{x^2}dx = \int \cot^2 \theta \ d\theta = -\cot\theta - \theta + C$$
Returning to the original variable, $$\int \frac{\sqrt{9 - x^2}}{x^2}dx = -\frac {\sqrt{9 - x^2}}{x} - \sin^{-1}\left(\frac{x}{3}\right) + C$$
I don't understand why $\sqrt{9-x^2} = 3|\cos\theta| = 3\cos\theta \,$ instead of $\sqrt{9-x^2} = |3||\cos\theta| = |3|\cos\theta$. I feel like I have problems understanding this because I am not sure what is the purpose of the absolute value signs in this case, are they to indicate that, for example, $|\cos\theta| = \pm\cos\theta$? If that's the case, why do we choose $3$ to be positive instead of negative?
|
Your book is being very careful about the fact that $\sqrt{1-\sin^2\theta} = \cos\theta$ is only true when $\cos \theta \geq 0$. Of course, in this problem it doesn't matter, since $-\frac{\pi}{2} < \theta \leq \frac{\pi}{2}$.
Note that you could just as well have chosen the substitution $x = -3\sin \theta$. It is a good exercise to check that this gives you the same answer as $x=3\sin \theta$.
|
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|
Show that $\lim_{n\rightarrow \infty}(\sqrt[3]{n+\sqrt{n}}-\sqrt[3]{n})=0$ I am asked to prove this statement $^{*}$. I am trying now, but it is getting to small and tiny steps that I even loose my way. my steps are as follows:
$$\lim_{n\rightarrow \infty}(\sqrt[3]{n+\sqrt{n}}-\sqrt[3]{n})=0^{*}$$
$\lim_{n\rightarrow \infty}(\sqrt[3]{n+\sqrt{n}}-\sqrt[3]{n})=\dfrac{(\sqrt[3]{n+\sqrt{n}}-\sqrt[3]{n}) \cdot (\sqrt[3]{n+\sqrt{n}}+\sqrt[3]{n})}{(\sqrt[3]{n+\sqrt{n}}+\sqrt[3]{n})}=\dfrac{(\sqrt[3]{n+\sqrt{n}})^2-(\sqrt[3]{n})^2}{(\sqrt[3]{n+\sqrt{n}}+\sqrt[3]{n})}=\dfrac{\sqrt[3]{n^2+2n\sqrt{n}+n}}{\sqrt[3]{n+\sqrt{n}}+\sqrt[3]{n}}=\dfrac{(n+\sqrt{n})^{\frac{2}{3}}-n^{\frac{2}{3}}}{(n+\sqrt{n})^{\frac{1}{3}}+n^{\frac{1}{3}}}= .. help = 0$ $$if \quad n\rightarrow \infty$$
|
What Dennis wrote. Or,
$$
\sqrt[3]{n+\sqrt{n}}-\sqrt[3]{n}=C\int_n^{n+\sqrt{n}}\frac{\mathrm dt}{t^{2/3}}\leqslant C\int_n^{n+\sqrt{n}}\frac{\mathrm dt}{C'n^{2/3}}=C''n^{1/2-2/3}\to0,
$$
because $1/2\lt2/3$. Likewise, for every positive $A$ and $B$,
$$
\sqrt[A]{n+\sqrt[B]{n}}-\sqrt[A]{n}\to0\iff\frac1A+\frac1B\lt1.
$$
|
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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.