Interpolation polynomial 
Consider the following table of values for a function $j_0(x)$:
$\begin{array}{c|ccccc} x & \delta_0(x)
 \\  \hline  0.0&1.00000\\0.1&.99833\\0.2&.99335\\0.3&.98507\\0.4&.97355\\0.5&.95885\\0.6&.94107\\0.7&.92031\\0.8&.89670\\0.9&.87036\\1.0&.84147\\1.1&.81019\\1.2&.77670\\1.3&.74120 
 \end{array}$
What should be the maximum degree of polynomial interpolation used
  with the table?

I know that I must use the forward difference table so that I can detect the influence of the rounding errors. From that, do I find the polynomial by using $$f[x_0,...,x_n]=\frac{\triangle^nf(x_0)}{n!h^n}$$ where $\triangle f(x)=f(x+h)-f(x)$ and $h$ is the step length? Or do I use newton divided difference? Confused on how I can solve this.
 A: The data is only given to 5 decimal places. So, we can assume that the data points might be in error by as much as 0.000005. So any curve that deviates from the data points by less than this would be fine. If you plot the data (in Excel, for example) you will see that the curve gently slopes downwards. I would guess that a curve of degree 5 or 6 would fit the points with an error of less than 0.000005. As far as I know, there is no good way to predict the error, other than constructing the curve and measuring. To construct curves, use least-squares fitting: http://mathworld.wolfram.com/LeastSquaresFittingPolynomial.html
To use the Wolfram formulae: In your case, $n=14$, $\{x_1, x_2, \ldots, x_n\} = \{0.0, 0.1, \ldots, 1.3\}$, and $\{y_1, y_2, \ldots, y_n\} = \{1.00000, 0.99833, \ldots, 0.74120\}$. But you can probably find some software to do the fitting for you. For example, you can use the "Fit" function in Mathematica. http://reference.wolfram.com/mathematica/ref/Fit.html
You can do it in Excel, too. Look up the "Add Trendline" function. The function $y = -0.0008x^5 + 0.0096x^4 - 0.0009x^3 - 0.1664x^2 - 0.00002x + 1$ gives a maximum error of $0.00003$.
