Question about expanding a function to first order

[hgandh]

If we have a function $f(x+\Delta x)$ where $\Delta x << x$, is it valid to approximate this as:
$$f(x + \Delta x) \approx f(x) + f'(x)\Delta x$$
even if $\Delta x$ is not necessarily small? If not, what is the valid expansion to first order?

 

[fresh_42]

If $\Delta x$ is big, then $f'(x)$ and $\dfrac{f(x+\Delta x)-f(x)}{\Delta x}$ can be very different, regardless how small $\Delta x$ versus $x$ is. E.g. $x$ could be positive and $\Delta x$ a large negative number, i.e. the function values can be quite different.

 

[hgandh]

If $\Delta x$ is big, then $f'(x)$ and $\dfrac{f(x+\Delta x)-f(x)}{\Delta x}$ can be very different, regardless how small $\Delta x$ versus $x$ is. E.g. $x$ could be positive and $\Delta x$ a large negative number, i.e. the function values can be quite different.

If I restrict it to positive values only, would this become valid then?

 

[PeroK]

If I restrict it to positive values only, would this become valid then?

No. To use a first order approximation, $\Delta x$ must be "small" in some sense.

You can try some examples, like $f(x)=x^2$.

 

[PeroK]

PS another perhaps even more illustrative example would be $\sin x$.

 

[WWGD]

Exceptions would be linear, constant functions where expansion would be valid everywhere.

 

[RPinPA]

If we have a function $f(x+\Delta x)$ where $\Delta x << x$, is it valid to approximate this as:
$$f(x + \Delta x) \approx f(x) + f'(x)\Delta x$$
even if $\Delta x$ is not necessarily small? If not, what is the valid expansion to first order?

You can answer this yourself. Draw a curve, any curve. That's $f(x)$. Draw the tangent line at any point $x_0$. That's the line $f(x_0) + f'(x_0) \Delta x$. How far is the line from the curve? Anywhere it's "close enough", whatever that means to you, is a place where the approximation is "good enough".

Intuitively you can see that if the curve bends rapidly away from the tangent line, then the error gets big quickly, whereas if the curve is relatively flat and stays close to the line, then the approximation is pretty good over a larger range.

Taylor's Theorem expresses that rigorously with the error term. If you stop at the first derivative, the amount of error depends on the second derivative, i.e. how fast the slope of the curve changes.

 

[WWGD]

You can answer this yourself. Draw a curve, any curve. That's $f(x)$. Draw the tangent line at any point $x_0$. That's the line $f(x_0) + f'(x_0) \Delta x$. How far is the line from the curve? Anywhere it's "close enough", whatever that means to you, is a place where the approximation is "good enough".

Intuitively you can see that if the curve bends rapidly away from the tangent line, then the error gets big quickly, whereas if the curve is relatively flat and stays close to the line, then the approximation is pretty good over a larger range.

Taylor's Theorem expresses that rigorously with the error term. If you stop at the first derivative, the amount of error depends on the second derivative, i.e. how fast the slope of the curve changes.

Interesting. This gives a formal proof that a linear function gives a perfect approximation, since f"(x)==0 for f(x)=mx+b.