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According to Wiki on Linear Approximation

Given a twice continuously differentiable function $f$ of one real variable, Taylor's theorem for the case $n = 1$ states that

$$f(x) = f(a) + f'(a)(x - a) + R_2$$

I'm not really dealing with Taylor polynomials yet, more learning calc and linear approximation currently. I'm wondering whether the condition for being twice continuously differentiable is required to apply linear approximation, and if so, why.

Ben G
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1 Answers1

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To understand what it means. Take any function $f:E\rightarrow \mathbb{R}$ and $x_0\in E$. Then consider $$R(x) = f(x)-f(x_0)-f'(x_0)(x-x_0)$$ You can always do that (and rearrange to get your expression), the thing that matters is exactly how that function $R(x)$ behaves when ${x\rightarrow x_0}$ (is it a good linear approximation or is it completely useless?)

For example, if $f$ is a differentiable function it means that. $$\lim_{h\rightarrow 0} \lVert R_h\rVert = \lim_{h\rightarrow 0} \lVert\frac{f(x_0+h)-f(x_0)}{h} - f'(x_0)\rVert = 0$$ So you already have that $f(x) = f(x_0) + f'(x_0) h +R_h$ and now you know that if f is differentiable it is a good approximation, how good? Well $\lim_{h\rightarrow 0} R_h = 0$. Perhaps you can do even better..

The extra differentiability conditions you mentioned give you more information with regards as to how the difference $R_h$ behaves (what aitor has commented)

ALakatos
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  • Is it fair to say I'm unlikely to run into a function that is only once differentiable? It seems like they have to be elaborately constructed https://math.stackexchange.com/questions/1414729/only-once-differentiable – Ben G Jan 11 '22 at 23:11
  • Also a more core question: suppose you don't care about $R$, can you still apply linear approximation? Or it's impossible unless twice differentiable? – Ben G Jan 11 '22 at 23:16
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    Actually NO. From an advance point of view, it can be shown that nowhere-differentiable functions make a much larger set than the set of functions that have, at least, derivative at one point. It is said that the set of nowhere-differentiable functions is of the second category and that the st of differentiable functions at least in one point is of the first category. Check "Measure and Category" by John Oxtoby if you are interested – juan zaragoza Jan 11 '22 at 23:19
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    @juanzaragoza Correct, of course, and an interesting result. That said, bgcode, I think it IS fair to say you generally won't run into functions that are only once differentiable. If we were to pick a continuous function "at random", it will almost certainly be nowhere differentiable, but when we run into functions in day-to-day life, they tend to be nice and smooth. – Theo Bendit Jan 11 '22 at 23:25
  • Thanks, note that "run into" was from perspective of a math learner rather than a picker from the function universe. Core remaining question: can you forget about R and get some kind of approximation on once differentiable functions, or any linear approximation is impossible / pointless? – Ben G Jan 11 '22 at 23:27
  • @bgcode Regarding your core question, you always care about $R$ (i.e. the error) in approximation, to some extent. It is a measure of "how good" your approximation is. If you asked me to approximate $\pi$ and I said something stupid like $-100$, it would technically be an approximation, just an approximation with stupidly large error that makes it unsuitable for most purposes. If I said $\frac{22}{7}$, then the error is much smaller, and it's suitable for many purposes. If you don't have any idea of how small your error is, you're flying blind! – Theo Bendit Jan 11 '22 at 23:31
  • So your error could be infinitely large if you can't take the second derivative and attempt to approximate? – Ben G Jan 11 '22 at 23:40
  • No real number is "infinitely large". – Lee Mosher Jan 12 '22 at 00:06
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    @bgcode You could say that the error could be "unboundedly" large. You have no control over how different the approximation is from the true number itself. This is still true if the function is twice differentiable, if you have no idea how large the second derivative could get. The $f''(b)$ term in the error could potentially be quite large if $f''$ is not well behaved between your point $x$ and the point $a$; Taylor's theorem just guarantees us a point $b$ in between $x$ and $a$ such that the error is $f''(b)(x - a)^2/2$, so if $f''$ blows up, then so might $f''(b)$. – Theo Bendit Jan 12 '22 at 00:11