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Use Matlab (or Python) to fit a cubic polynomial to data, and determine where the x-axis is crossed.

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This question tests polynomial curve fitting and root solving.

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By fitting a cubic polynomial to the following data:

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
t0123456789
Y$\\var{Y0}$$\\var{Y1}$$\\var{Y2}$$\\var{Y3}$$\\var{Y4}$$\\var{Y5}$$\\var{Y6}$$\\var{Y7}$$\\var{Y8}$$\\var{Y9}$
\n
    \n
  1. The value of Y when t=10 is:
    2. \n
  2. When Y=0, the value of x is:
    3. \n
\n

In Matlab, for example:

\n
t = 0:9;
Y = [{Y0},{Y1},{Y2},{Y3},{Y4},{Y5},{Y6},{Y7},{Y8},{Y9}];
coeffs = polyfit(t, Y, 3);   % fit a cubic polynomial to the data; in Python, use NumPy and np.polyfit(), etc.
y = @(t) polyval(coeffs, t); % define a function based on our curve fit
\n

Q1. y(10) = $\\var{Y10}$

\n
roots(coeffs); % find where the polynomial is zero... here there is only one real solution
\n

Q2. y($\\var{c}$) = 0

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The one real root of the polymonial.

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Value of Y when t=0.

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Value of Y when t=1.

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Value of Y when t=2.

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Value of Y when t=3.

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Value of Y when t=4.

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Value of Y when t=5.

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Value of Y when t=6.

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Value of Y when t=7.

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Value of Y when t=8.

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Value of Y when t=9.

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Value of Y when t=10.

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By fitting a cubic polynomial to the following data:

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
t0123456789
Y$\\var{Y0}$$\\var{Y1}$$\\var{Y2}$$\\var{Y3}$$\\var{Y4}$$\\var{Y5}$$\\var{Y6}$$\\var{Y7}$$\\var{Y8}$$\\var{Y9}$
\n
    \n
  1. The value of Y when t=10 is: [[0]]
    2. \n
  2. When Y=0, the value of t is: [[1]]
    3. \n
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