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Given the table of monthly rainfall below, how do you model it with a sinusoidal function of the form #f(x) = A sin(omega x - phi) + beta# where #x# is the month number #1-12# ?

George C.
Featured 5 months ago

Here's a rough method...

Explanation:

Here's one naive and slightly long-winded approach, which may or may not be the sort of thing the posers of the question had in mind...

Given:

$\left[\begin{matrix}\text{January" & 1 & 3.91 \\ "February" & "2" & 4.36 \\ "March" & 3 & 5.31 \\ "April" & 4 & 6.21 \\ "May" & 5 & 7.02 \\ "June" & 6 & 7.84 \\ "July" & 7 & 8.19 \\ "August" & 8 & 8.06 \\ "September" & 9 & 7.41 \\ "October" & 10 & 6.30 \\ "November" & 11 & 5.21 \\ "December} & 12 & 4.28\end{matrix}\right]$

We want to match this to:

$f \left(x\right) = A \sin \left(\omega x - \phi\right) + \beta$

where $A$ is the amplitude, $\phi$ is the phase and $\beta$ the offset.

The natural period of $\sin \theta$ is $2 \pi$, whereas the natural period of this data is $12$ (in $x$).

So $\omega = \frac{\pi}{6}$.

Note that $\cos \theta$ is $\frac{\pi}{2}$ ahead of $\sin \theta$, so pairing values with a $3$ month offset should give us the coordinates of points roughly on a circle:

$\left(6.21 , 3.91\right)$, $\left(7.02 , 4.36\right)$, $\left(7.84 , 5.31\right)$,

$\left(8.19 , 6.21\right)$, $\left(8.06 , 7.02\right)$, $\left(7.41 , 7.84\right)$,

$\left(6.30 , 8.19\right)$, $\left(5.21 , 8.06\right)$, $\left(4.28 , 7.41\right)$,

$\left(3.91 , 6.30\right)$, $\left(4.36 , 5.21\right)$, $\left(5.31 , 4.28\right)$

The centre of the circle is $\left(m , m\right)$ where $m$ is the mean of the values...

$m = \frac{1}{12} \left(3.91 + 4.36 + 5.31 + 6.21 + 7.02 + 7.84 + 8.19 + 8.06 + 7.41 + 6.30 + 5.21 + 4.28\right) = 6.175$

If all of the $12$ points lay exactly on the circle, then the distance of all of them from $\left(6.175 , 6.175\right)$ would be the same. Let's find the actual distances to $3$ significant figures, then take the average:

The distance between $\left({x}_{1} , {y}_{1}\right)$ and $\left({x}_{2} , {y}_{2}\right)$ is given by the formula:

$\sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}$

Putting $\left({x}_{1} , {y}_{1}\right) = \left(6.175 , 6.175\right)$ and trying each of the $12$ points listed above as $\left({x}_{2} , {y}_{2}\right)$, we find distances:

$2.27$, $2.00$, $1.88$, $2.02$, $2.07$, $2.07$, $2.02$, $2.12$, $2.26$, $2.27$, $2.06$, $2.08$

with mean: $2.09$

So far, we have determined $A = 2.09$, $\omega = \frac{\pi}{6}$, $\beta = 6.175$

Notice that $\beta - A = 6.175 - 2.09 = 4.085$ which is already larger than the January value.

So we might as well approximate the minimum of the sinusoid at $x = 1$, so give $\phi = \frac{\pi}{6} + \frac{\pi}{2} = \frac{2 \pi}{3}$

So our formula is:

$f \left(x\right) = A \sin \left(\omega x - \phi\right) + \beta = 2.09 \sin \left(\frac{\pi}{6} x - \frac{2 \pi}{3}\right) + 6.175$

which results in monthly values:

$4.09$, $4.37$, $5.13$, $6.17$, $7.22$, $7.98$,

$8.27$, $7.98$, $7.22$, $6.17$, $5.13$, $4.37$

graph{(y-(2.09 sin(pi x / 6-2pi/3) + 6.175))((x-1)^2+(y-3.91)^2-0.03)((x-2)^2+(y-4.36)^2-0.03)((x-3)^2+(y-5.31)^2-0.03)((x-4)^2+(y-6.21)^2-0.03)((x-5)^2+(y-7.02)^2-0.03)((x-6)^2+(y-7.84)^2-0.03)((x-7)^2+(y-8.19)^2-0.03)((x-8)^2+(y-8.06)^2-0.03)((x-9)^2+(y-7.41)^2-0.03)((x-10)^2+(y-6.30)^2-0.03)((x-11)^2+(y-5.21)^2-0.03)((x-12)^2+(y-4.28)^2-0.03) = 0 [-0.85, 19.15, 1.56, 11.56]}

Question #4b5ea

Douglas K.
Featured 5 months ago

${x}^{2} - 2 x y + {y}^{2} + 2 x + 2 y - 1 = 0$

Explanation:

Given:

focus: $\left(0 , 0\right)$
directrix: $x + y = 1$

Rewrite the equation for the directrix as:

$x + y - 1 = 0$

This is done to that it fits the formula for the distance from a point to a line:

$d = | a x + b y + c \frac{|}{\sqrt{{a}^{2} + {b}^{2}}} \text{ [1]}$

where $\left(x , y\right)$ is a point on the parabola $a = 1 , b = 1 , \mathmr{and} c = - 1$

$d = | x + y - 1 \frac{|}{\sqrt{{1}^{2} + {1}^{2}}} \text{ [1.1]}$

$d = | x + y - 1 \frac{|}{\sqrt{2}} \text{ [1.2]}$

The distance from the focus, $\left(0 , 0\right)$, to the a point $\left(x , y\right)$ on the parabola is:

$d = \sqrt{{\left(x - 0\right)}^{2} + {\left(y - 0\right)}^{2}} \text{ [2]}$

$d = \sqrt{{x}^{2} + {y}^{2}} \text{ [2.1]}$

A parabola is the locus of points that are equidistant from its focus and its directrix, therefore, we can derive the equation of the parabola by setting the right side of equation [1.2] equal to the right side of equation [2.1]:

$| x + y - 1 \frac{|}{\sqrt{2}} = \sqrt{{x}^{2} + {y}^{2}}$

Multiply both sides by $\sqrt{2}$

$| x + y - 1 | = \sqrt{2 {x}^{2} + 2 {y}^{2}}$

Square both sides:

${\left(x + y - 1\right)}^{2} = 2 {x}^{2} + 2 {y}^{2}$

Expand the square:

${x}^{2} + 2 x y - 2 x + {y}^{2} - 2 y + 1 = 2 {x}^{2} + 2 {y}^{2}$

Combine like terms and write the equation in the General Cartesian form of a conic section , $A {x}^{2} + B x y + C {y}^{2} + D x + E y + F = 0$:

${x}^{2} - 2 x y + {y}^{2} + 2 x + 2 y - 1 = 0$

Here is a graph of the parabola, the focus and the directrix:

How can you prove that #log_10 2# is irrational ?

Rhys
Featured 5 months ago

We can prove this by contradiction;

Explanation:

if $x = \mathbb{Q}$ then $x = \frac{p}{q}$ for $p , q \in \mathbb{Z}$

so hence let, ${\log}_{10} 2 = \frac{p}{q}$
then $2 = {10}^{\frac{p}{q}}$

so hence ${2}^{q} = {10}^{p}$

$\implies {2}^{q} = {5}^{p} {2}^{p}$

$\implies {2}^{q - p} = {5}^{p}$

we know ${\log}_{10} 1 < {\log}_{10} 2 < {\log}_{10} 10$

so hence $0 < {\log}_{10} 2 < 1$

so hence $0 < \frac{p}{q} < 1$

$p < q$

so $q - p > 0$ and let $m = q - p$

But we can see that ${2}^{m}$ is even and ${5}^{p}$ is odd, hence there are no values for $p \mathmr{and} q$ that satisify this

$\implies$ Hence proven by contradiction

How do you multiply #((1, -3, 2), (2, 1, -3), (4, -3, -1))# and #((2, 1, -1, -2), (3, -2, -1, -1), (2, -5, -1, 0))#?

George C.
Featured 4 months ago

$\left(\begin{matrix}1 & - 3 & 2 \\ 2 & 1 & - 3 \\ 4 & - 3 & - 1\end{matrix}\right) \left(\begin{matrix}2 & 1 & - 1 & - 2 \\ 3 & - 2 & - 1 & - 1 \\ 2 & - 5 & - 1 & 0\end{matrix}\right) = \left(\begin{matrix}- 3 & - 3 & 0 & 1 \\ 1 & 15 & 0 & - 5 \\ - 3 & 15 & 0 & - 5\end{matrix}\right)$

Explanation:

For each row of the first matrix and column of the second, sum the products of the three corresponding terms...

#((color(red)(1), color(red)(-3), color(red)(2)), (2, 1, -3), (4, -3, -1))((color(blue)(2), 1, -1, -2),(color(blue)(3), -2, -1, -1),(color(blue)(2), -5, -1, 0)) = ((color(purple)(-3), ?, ?, ?),(?, ?, ?, ?), (?, ?, ?, ?))#

$\left(\textcolor{red}{1}\right) \left(\textcolor{b l u e}{2}\right) + \left(\textcolor{red}{- 3}\right) \left(\textcolor{b l u e}{3}\right) + \left(\textcolor{red}{2}\right) \left(\textcolor{b l u e}{2}\right) = 2 - 9 + 4 = \textcolor{p u r p \le}{- 3}$

#((color(red)(1), color(red)(-3), color(red)(2)), (2, 1, -3), (4, -3, -1))((2, color(blue)(1), -1, -2),(3, color(blue)(-2), -1, -1),(2, color(blue)(-5), -1, 0)) = ((-3, color(purple)(-3), ?, ?),(?, ?, ?, ?), (?, ?, ?, ?))#

$\left(\textcolor{red}{1}\right) \left(\textcolor{b l u e}{1}\right) + \left(\textcolor{red}{- 3}\right) \left(\textcolor{b l u e}{- 2}\right) + \left(\textcolor{red}{2}\right) \left(\textcolor{b l u e}{- 5}\right) = 1 + 6 - 10 = \textcolor{p u r p \le}{- 3}$

#dot dot dot#

#((1, -3, 2), (color(red)(2), color(red)(1), color(red)(-3)), (4, -3, -1))((color(blue)(2), 1, -1, -2),(color(blue)(3), -2, -1, -1),(color(blue)(2), -5, -1, 0)) = ((-3, -3, 0, 1),(color(purple)(1), ?, ?, ?), (?, ?, ?, ?))#

$\left(\textcolor{red}{2}\right) \left(\textcolor{b l u e}{2}\right) + \left(\textcolor{red}{1}\right) \left(\textcolor{b l u e}{3}\right) + \left(\textcolor{red}{- 3}\right) \left(\textcolor{b l u e}{2}\right) = 4 + 3 - 6 = \textcolor{p u r p \le}{1}$

#dot dot dot#

$\left(\begin{matrix}1 & - 3 & 2 \\ 2 & 1 & - 3 \\ \textcolor{red}{4} & \textcolor{red}{- 3} & \textcolor{red}{- 1}\end{matrix}\right) \left(\begin{matrix}2 & 1 & - 1 & \textcolor{b l u e}{- 2} \\ 3 & - 2 & - 1 & \textcolor{b l u e}{- 1} \\ 2 & - 5 & - 1 & \textcolor{b l u e}{0}\end{matrix}\right) = \left(\begin{matrix}- 3 & - 3 & 0 & 1 \\ 1 & 15 & 0 & - 5 \\ - 3 & 15 & 0 & \textcolor{p u r p \le}{- 5}\end{matrix}\right)$

$\left(\textcolor{red}{4}\right) \left(\textcolor{b l u e}{- 2}\right) + \left(\textcolor{red}{- 3}\right) \left(\textcolor{b l u e}{- 1}\right) + \left(\textcolor{red}{- 1}\right) \left(\textcolor{b l u e}{0}\right) = - 8 + 3 + 0 = \textcolor{p u r p \le}{- 5}$

How to do questions b?

Somebody N.
Featured 3 months ago

$\boldsymbol{Y} = \left[\begin{matrix}- \frac{11}{16} & \frac{17}{16} \\ - \frac{1}{4} & \frac{3}{4}\end{matrix}\right]$

Explanation:

First find $\boldsymbol{{A}^{-} 1}$

The easiest way to find the inverse of $\boldsymbol{A}$ is to find the determinant of $\boldsymbol{A}$:

This is just:

$\left(3 \times 6\right) - \left(1 \times 2\right) = 16$

Next switch the elements on the leading diagonal of $\boldsymbol{A}$ and change the signs of the elements on the non-leading diagonal of $\boldsymbol{A}$

So you should have:

$\left[\begin{matrix}6 & - 2 \\ - 1 & 3\end{matrix}\right]$

Divide each element by the determinant $\boldsymbol{16}$:

$\left[\begin{matrix}\frac{6}{16} & - \frac{2}{16} \\ - \frac{1}{16} & \frac{3}{16}\end{matrix}\right] = \left[\begin{matrix}\frac{3}{8} & - \frac{1}{8} \\ - \frac{1}{16} & \frac{3}{16}\end{matrix}\right]$

$\boldsymbol{{A}^{-} 1} = \left[\begin{matrix}\frac{3}{8} & - \frac{1}{8} \\ - \frac{1}{16} & \frac{3}{16}\end{matrix}\right]$

Now:

$\boldsymbol{Y A} + \boldsymbol{B} = \boldsymbol{C}$

$\boldsymbol{Y A} = \boldsymbol{C - B}$

Using $\boldsymbol{{A}^{-} 1}$

$\boldsymbol{Y A {A}^{-} 1} = \boldsymbol{\left(C - B\right) {A}^{-} 1}$

$\boldsymbol{Y} = \boldsymbol{\left(C - B\right) {A}^{-} 1}$

Note we are post multiplying on both sides. This is important as Matrix multiplication is non-commutative.

i.e.

$\boldsymbol{A B} \ne \boldsymbol{B A}$ ( In general )

$\boldsymbol{C - B} = \left[\begin{matrix}3 & 4 \\ 2 & 6\end{matrix}\right] - \left[\begin{matrix}4 & - 1 \\ 2 & 2\end{matrix}\right] = \left[\begin{matrix}- 1 & 5 \\ 0 & 4\end{matrix}\right]$

$\therefore$

$\boldsymbol{Y} = \left[\begin{matrix}- 1 & 5 \\ 0 & 4\end{matrix}\right] \left[\begin{matrix}\frac{3}{8} & - \frac{1}{8} \\ - \frac{1}{16} & \frac{3}{16}\end{matrix}\right] = \left[\begin{matrix}- \frac{11}{16} & \frac{17}{16} \\ - \frac{1}{4} & \frac{3}{4}\end{matrix}\right]$

$\boldsymbol{Y} = \left[\begin{matrix}- \frac{11}{16} & \frac{17}{16} \\ - \frac{1}{4} & \frac{3}{4}\end{matrix}\right]$

What is the sum of the first #n# terms of the series #1/2+3/4+5/8+...# ?

Shwetank Mauria
Featured 3 months ago

${S}_{n} = 3 - \frac{2 n + 3}{2} ^ n$

Explanation:

Let ${S}_{n} = \frac{1}{2} + \frac{3}{4} + \frac{5}{8} + \ldots \ldots \ldots \ldots \ldots \ldots \text{n terms}$

What is ${n}^{t h}$ term of the series? Well numerators are in arithmatic sequence and numerator of ${n}^{t h}$ term is $2 n - 1$ and denominator is in geometric series with ${n}^{t h}$ term being ${2}^{n}$. Hence ${n}^{t h}$ term of given series is $\frac{2 n - 1}{2} ^ n$ and series is

${S}_{n} = \frac{1}{2} + \frac{3}{4} + \frac{5}{8} + \frac{7}{16} + \ldots \ldots \ldots \ldots \ldots \ldots + \frac{2 n - 1}{2} ^ n$ and then

$\frac{1}{2} {S}_{n} = \frac{1}{4} + \frac{3}{8} + \frac{5}{16} + \ldots \ldots \ldots \ldots \ldots \ldots + \frac{2 n - 1}{2} ^ \left(n + 1\right)$

Subtracting latter from former, we get

${S}_{n} - \frac{1}{2} {S}_{n} = \frac{1}{2} + \frac{2}{4} + \frac{2}{8} + \ldots . + \frac{2}{2} ^ n - \frac{2 n - 1}{2} ^ \left(n + 1\right)$

or $\frac{1}{2} {S}_{n} = \frac{1}{2} + \frac{2}{4} \left[1 + \frac{1}{2} + \ldots . + \frac{1}{2} ^ \left(n - 2\right)\right] - \frac{2 n - 1}{2} ^ \left(n + 1\right)$

= $\frac{1}{2} + \frac{1}{2} \frac{1 - \frac{1}{2} ^ \left(n - 1\right)}{1 - \frac{1}{2}} - \frac{2 n - 1}{2} ^ \left(n + 1\right)$

= $\frac{1}{2} + 1 - \frac{1}{2} ^ \left(n - 1\right) - \frac{2 n - 1}{2} ^ \left(n + 1\right)$

= $\frac{3}{2} - \frac{4 + 2 n - 1}{2} ^ \left(n + 1\right)$

= $\frac{3}{2} - \frac{2 n + 3}{2} ^ \left(n + 1\right)$

and ${S}_{n} = 3 - \frac{2 n + 3}{2} ^ n$

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