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Draw a scatter diagram of the data for one period. Find the sinusoidal function of the form y=A sin(ωx-ϕ)+β that fits the data. Show your work.

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Nov 14, 2017

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]}

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SCooke Share
Jan 26, 2018

$y = 5 \sin \left(\omega x + \phi\right) + 8$ General Form
$y = 5 \sin \left(3.996 x - 31.28\right) + 8$ Form fit (loosely) to data.

Explanation:

This looks more complicated than it is. Break it down into the basic components. You are given the equation form (GOOD - often we have to figure that out on our own!) that we want to fit the data to. Nevermind the complicated form, basically , it is a sine function, modified by an amplitude (A) and an offset (B). Everything inside the sin() doesn't matter, as whatever they are must reduce to some value from which the sine is calculated.

Thus, we really just need to "fit" the equation:
$y = A \sin \left(\omega\right) + B$

The only observed value it the rainfall, so the only "x" metric is the month. We should end up with a cyclical curve showing the expected rainfall as a function of the month number, 1-12.

A sine function only has values between $- 1 \mathmr{and} + 1$, so to get a result of 8.19 the offset "B" for the amplitude must be at least that much from +1, or 7.19. Practically, and offset of 8 would allow for possible higher rainfalls. Set $B = 8$.

Similarly, the data range is 3.91 - 8.19, or $\Delta 4.28$ units. To fit into the $- 1 \mathmr{and} + 1$ range of a sine function, the amplitude multiplier "A" must be at least this value. Again, for practical use we would set it at 5 to allow for possible higher values and to keep the curve on the graph nicely. Set $A = 5$.

NOW we have the basic equation:
$y = 5 \sin \left(\omega x + \phi\right) + 8$

Given a value of y and a month number (1-12) calculate the multipliers $\omega \mathmr{and} \phi$.

For May this would be:
$7.02 = 5 \sin \left(\omega 5 + \phi\right) + 8$

$\frac{7.02 - 8}{5} = \sin \left(\omega 5 + \phi\right)$
$- 0.196 = \sin \left(\omega 5 + \phi\right)$
$\omega 5 + \phi = - 11.3$ ; $\omega \times 5 = - 11.3 - \phi$
$\omega = \frac{- 11.3 - \phi}{5}$

For August the same process gives:
$8.06 = 5 \sin \left(\omega 8 + \phi\right) + 8$

$\frac{8.06 - 8}{5} = \sin \left(\omega 8 + \phi\right)$
$0.012 = \sin \left(\omega 8 + \phi\right)$
$\omega 8 + \phi = 0.688$
$\omega = \frac{0.688 - \phi}{8}$

Set these equal ($\omega$ is the same) to find $\phi$:
$\frac{0.688 - \phi}{8} = \frac{- 11.3 - \phi}{5}$
$3.44 - 5 \phi = - 90.4 - 8 \phi$
$93.84 = - 3 \phi$; $\phi = - 31.28$:

Omega is then $\omega = \frac{0.688 + 31.28}{8} = 3.996$

The final equation is thus:
$y = 5 \sin \left(3.996 x - 31.28\right) + 8$

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Nov 14, 2017

See below.

Explanation:

Considering

$E \left(A , \omega , \phi , \beta\right) = {\sum}_{k = 1}^{12} {\left({y}_{k} - A \sin \left(\omega {x}_{k} - \phi\right) - \beta\right)}^{2} + {\omega}^{2}$ we have

$\frac{\partial E}{\partial A} = 2 {\sum}_{k = 1}^{12} \sin \left(\omega {x}_{k} - \phi\right) \left({y}_{k} - A \sin \left(\omega {x}_{k} - \phi\right) - \beta\right) = 0$

$\frac{\partial E}{\partial \omega} = - 2 {\sum}_{k = 1}^{12} A {x}_{k} \cos \left(\omega {x}_{k} - \phi\right) \left({y}_{k} - A \sin \left(\omega {x}_{k} - \phi\right) - \beta\right) + 2 \omega = 0$

$\frac{\partial E}{\partial \phi} = 2 {\sum}_{k = 1}^{12} A \cos \left(\omega {x}_{k} - \phi\right) \left({y}_{k} - A \sin \left(\omega {x}_{k} - \phi\right) - \beta\right) = 0$

$\frac{\partial E}{\partial \beta} = - 2 {\sum}_{k = 1}^{12} \left({y}_{k} - A \sin \left(\omega {x}_{k} - \phi\right) - \beta\right) = 0$

Solving this system of nonlinear equations with a numerical method like Newton-Raphson's we get

$A = 2.2 , \omega = - 0.4784 , \phi = 20.189 , \beta = 5.975$

Attached a plot solving the fitting

NOTE:

The term ${\omega}^{2}$ was added to the cumulated square error $E$ to avoid aliasing problems.

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