Taylor series expansion or polynomial regression which is the correct term here for finding an analytical estimate for an unknown function?

Question

I have a question regarding the Taylor expansion of an unknown function and I would be tanksful to have your comments on that.

Suppose we want to find an analytical estimate for an unknown function. The available information for this function is; its exact value at $x_0$ $(f_0)$ and first derivative at $x_0$ $(f_0')$, and its exact value at $x=x_1$ $(f_1)$ which is far from $x_0$. This means that we only need to find the second derivative through solving $$f_1 = f_0 + f_0'(x_1-x_0) + \frac12 {f_0}''(x_1-x_0)^2$$ to form the truncated Taylor series for $f$.

My question is that whether the obtained quadratic function is the result of Taylor series expansion or polynomial regression? Is this a proper way of approximating a function considering that f"0 is found by matching the function value at $x_1$ which is far from $x_0$?

Answer 1

This is effectively a Taylor expansion built around $x=x_0$ but if you want a quadratic form for approximation, write $$f(x)\sim a+ bx +c x^2$$ and solve the three equations to get $$c=\frac {f_1-f_0}{(x_1-x_0)^2}-\frac {f'_0}{x_1-x_0}$$ $$b=\frac {f_1-f_0}{x_1-x_0}-c(x_0+x_1)$$ $$a=f_0-bx_0-c x_0^2$$

Let us try for $f(x)=e^x$ with $x_0=2$ and $x_1=4$ which give $$a=e^2 \left(e^2-4\right)\qquad b=-e^2 \left(e^2-4\right)\qquad c=\frac{1}{4} e^2 \left(e^2-3\right)$$ Computing for $x=3$, we get $$\frac{1}{4} e^2 \left(5+e^2\right) \sim 22.8859 $$ while $e^3=20.0855$.

For this case, it is not fantastic at all but it could be much better with a function growing slower than the exponential.

Since you have now the formulae for $(a,b,c)$, play with any function of your choice and, please, let me know.

Answer 2

Thanks for your answer Claude and sorry for the late response.

I thinks that for this case the quadratic polynomial and second-order Taylor expansion are identical as you can see for the both functions in the below figs.

Moreover, as it is clear the result is not accurate even locally when the Taylor series coefficients are calculated based on two different points, i.e. by knowing f(x0), f'(x0) and f(x1), as opposed to f(x0), f'(x0) and f"(x0). Obviously, this is much more pronounced for an exponential function.

To sum up, when estimating a function by Taylor expansion, the coefficients must be calculated from a single point.

In the following figs, x0=2 and x1=6. the Taylor expansions are either calculated based on f(2), f'(2) and f"(2) or f(2), f'(2) and f(6).