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I am reading up on Taylor approximation of a function and I'm trying to develop the intuition for the remainder, when approximating a function with $n^{th}$ degree polynomial which has a continuous $(n+1)^{th}$ derivate, given by $\frac{1}{n!}\int_{a}^{x} (x - t)^nf^{(n+1)}(t)dt$

My intuition of linear approximation is this: We used a constant first derivate to evaluate at x (since we approximate f at a). Hence, we have to use the information about rate of rate of change from the any point $ t \in (a,x)$ to compensate for this error. Specifically, the second derivative gives the difference between the first derivatives at two successive points and scales it over unit interval. Therefore, f''(t) corrects for error at t but introduces new error from (t, x) which is corrected with the same logic at the next point. Thus, the integral given above. Is this correct?

My reasoning is because if I begin the approximation using a constant function and reason that by using the rate of change at every point, a function can be reconstructed starting from any point. But if I try to use the above integral to compute the error in estimates for the constant function, it doesn't work because of the $(-t)$. Is there a formula to estimate the error including the constant case?

I understand the proof of the integral using integration by parts (and requirement of the continuity of $f^{(n+1)}(x)$ is to be able to use the first fundamental theorem).

Can you please help me fix my intuition of the integral?

Srini Vas
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1 Answers1

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Here is some intuition: Let $j_a^nf$ be the $n^{\rm th}$ Taylor polynomial of $f$ at $a$. If the $(n+1)^{\rm st}$ derivative of $f$ were identically zero then this polynomial $j_a^nf$ would produce the exact value of $f$ for all $x$, and the error $R(x):=f(x)-j_a^nf(x)$ would be identically zero. It follows that in the case of nonzero error the nonvanishing of $f^{(n+1)}(t)$ in certain points of the interval $[a,x]$ (when $x>a$) has to be the culprit. Linearity then would indicate that we have a formula of the form $$R_n(x)=\int_a^x w(t)f^{(n+1)}(t)\>dt$$ with a certain more or less "universal" weight function $w$. That this weight function has the simple form appearing in "Taylor's theorem with integral remainder" is due to the secret of partial integration, an intuitive explanation of which I still have to see.

Christian Blatter
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  • Thanks for the answer. To me the weight function was the obvious. Does the following makes sense? Suppose we work with a linear approximation, then we use the first derivate of f at a to compute P(x) where P is first degree polynomial approximation of f. As a result, we have incurred an error if first derivative is not a constant in (a, x). The error can be corrected with the second derivate because the difference quotient gives the difference between f'(x+h) and f'(x) and normalize it for a unit interval (by dividing by h). – Srini Vas Aug 26 '18 at 16:35
  • Thus, we scale this difference by (x - t), which is the distance to x from t (let t be the immediate point after a). This corrects for the error, arising from using the constant first derivate from a, at the point t. But introduces error in the estimation from (t, x] since we have used the constant difference of f'(t) - f'(a). This is corrected by the second derivate at t', the point immediately after t. This process is done at every point in [a,x] and thus the giving rise to the said integral. – Srini Vas Aug 26 '18 at 16:41