Newton polynomial interpolation degree

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It is asked to find a polynomial of adequate degree to estimate $\sqrt{1.035}$. The following table is given: Table

We know that 1.03 and 1.04 need to be used. From the divided differences table, we see that they are increasing, and we also have some zeros in the second-order divided differences.

I don't get what the author means by "adequate degree". For example, if we start in 1.02, we only can interpolate by a line. But I cannot decide if I should start in 1.01 or 1.

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The first order differences are almost constant here, they decrease very slowly. The simple linear interpolation between $1.03$ and $1.04$ is appropriate. It gives $$ \sqrt{1.035}\approx \frac{1.0149+1.0198}{2} = 1.01735 $$


Comparing this to $\sqrt{1.035} = 1.017349497\ldots$ confirms there is no need for higher order interpolation. The result has as many correct significant digits as the input $1.035$.

Another sign that higher degrees are not useful: quadratic interpolation gives a polynomial with leading coefficient about $10^{-12}$.