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In general, it holds that: $$f(x_0+\Delta x)\approx f(x_0)+f'(x_0)\cdot \Delta x$$ Clearly, an $e(\Delta x)$ error is made in the approximation.

Let us consider the equation $$f(x_0+\Delta x)-f(x_0)=f'(x_0)\cdot \Delta x + e(\Delta x)\cdot \Delta x$$ This equation tells us that the true increment of the function, which is a variable quantity dependent on $\Delta x$, is equal to the sum of the linear increment calculated on the tangent line and an error. It is not certain that $f'(x_0)\cdot \Delta x$ is a good approximation of $\Delta y$ in a neighborhood of $x_0$. However, it turns out that:

  • the error $e(\Delta x)\cdot \Delta x$ is infinitesimal for $\Delta x \to 0$.
  • the error $e(\Delta x)\cdot \Delta x$ is an infinitesimal of higher order than $\Delta x$.

Thus, there certainly exists a neighborhood of $x_0$ in which $e(\Delta x)\cdot \Delta x$ is less than the increment $\Delta x$ and therefore this ensures that the approximation is good since, for $ \Delta x$ small enough, the error is limited by the increment itself.

Is it a correct interpretation of differential approximation? If it is, is there an explicit expression for the error made?

Sigma Algebra
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1 Answers1

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Taylor's Theorem provides a bound on the error that for the linear approximation depends on the second derivative. In particular, if $|f''(t)| \le M$ for all $t$ between $x$ and $x+ \Delta x$ then the error in the linear approximation is at most $$ \frac{M (\Delta x )^2}{2}. $$

(If you had an "explicit expression for the error" you would not need the approximation.)

Ethan Bolker
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  • Thank you very much! Is the previous comment correct? Is it right that I write $e(\Delta x)\cdot \Delta x$ and not just $e(\Delta x)$? – Sigma Algebra Nov 02 '24 at 17:57
  • @SigmaAlgebra You can write the error term either way, as a function of $\Delta x$ or as a function to multiply $\Delta x$ by. That's just a matter of usefulness or taste. – Ethan Bolker Nov 02 '24 at 18:00
  • What is the most clear way in your opinion? :) – Sigma Algebra Nov 02 '24 at 18:03
  • I would write the error as a function of $x$ and $\Delta x$ rather than "factoring out" the $\Delta x$. The $(\Delta x)^2$ in Taylor's Theorem shows that the error is a "higher order infinitesimal". – Ethan Bolker Nov 02 '24 at 18:05
  • So, a better way is to write $f(x_0+\Delta x)-f(x_0)=f'(x_0)\cdot \Delta x+e(\Delta x)$ where $\lim_{\Delta x\to 0}\frac{e(\Delta x)}{\Delta x}=0$? – Sigma Algebra Nov 02 '24 at 18:09
  • Yes. But note that $e(x)$ is really $e(x, \Delta x)$. – Ethan Bolker Nov 02 '24 at 18:11
  • Okay, that's because the error depends on both the point and the increment. Thank you very much for the patience! – Sigma Algebra Nov 02 '24 at 18:13