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Is it possible for an elementary continuous and non-piecewise function to be linear on some part of the interval and nonlinear on another part of the interval?

For instance; a function has second derivative zero on [1,4] and after x = 4, it starts to curve. Also note that piecewiseness and infinitely many terms are not allowed.

artmyb
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    What do you call an elementary function ? –  Apr 23 '19 at 12:28
  • $ f_{t} = \frac{1}{t} \ln(1+e^{tx}) $ where you pick some big $t$. I'm not sure if that is ok. That function look" linear" to me for small x and "no-linear " to me when x is close to $0$. See: https://math.stackexchange.com/questions/3194279/smooth-approximation-of-fx-begincases0-textif-x0-x-textif-x-geq – user614287 Apr 23 '19 at 12:38
  • You rule out piecewise functions, but have a look at $\sqrt{x^2}$ or $\arcsin(\sin x)$... – Jean-Claude Arbaut Apr 23 '19 at 12:54
  • @ Jean-Claude Arbaut sorry, so I still don't understand why the function that I wrote is not good? – user614287 Apr 23 '19 at 13:19
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    This question may not be worded 100% formally, but is it really unclear what the OP is asking? – Ovi Apr 23 '19 at 22:48

2 Answers2

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You can write a piecewise function in terms of absolute value and/or max and min functions. So it should be possible.

Example: $f(x)=max(x×|x|,0)$

Using the formula $max(a,b)= \frac{a+b+|a-b|}{2}$ you can get an expression with absolute values

Miguel
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    The absolute value is piecewise linear and must be ruled out. –  Apr 23 '19 at 12:53
  • @YvesDaoust It's unclear in the question. The absolute value can be written $\sqrt{x^2}$, and it's not a piecewise definition. Then $\max(a,b)=\frac12(a+b+|a-b|)$. So you can get a quite "piecewise" function with only one expression involving only "elementary" functions. The concept does not look mathematically sound anyway (a function is not defined by an expression but by a set of couples $(x,f(x))$: the piecewiseness depends on the way the function is defined, and there may be several equivalent ways. – Jean-Claude Arbaut Apr 23 '19 at 12:57
  • @Jean-ClaudeArbaut: of course this is unclear. We shouldn't contribute until the OP has clarified. (Or close the question.) –  Apr 23 '19 at 12:57
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Miguel provided a good answer with the absolute value trick. But if you want to only use nice, smooth (elementary) functions which are differentiable everywhere, the answer is no.

All of the elementary functions are analytic (a term which I will define later), and the composition/sum/product/etc. of analytic functions is analytic, so any nice function that you come up with is analytic.

Now "analytic" means that if you give a point $p$ somewhere in their domain, you can give a power series $\sum a_k (x-p)^k$ that converges to your function in a small interval around $p$. Now let $p$ be the point where your function goes from being linear to being non-linear, and try to find the power series expansion around $p$. On one side, it has to be $a_0+a_1(x-p)+0+0(x-p)^2+0(x-p)^3+ \cdots$, but on the other side it has to be something else. This is impossible.

Ovi
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  • These discussions about analytic vs piecewise definition date back, I believe, to Euler or Lagrange. – Jean-Claude Arbaut Apr 23 '19 at 13:07
  • @Jean-ClaudeArbaut Hmm what is the discussion about analytic vs piecewise function? – Ovi Apr 23 '19 at 13:13
  • IIRW something I read about the history of mathematics, mathematicians were reluctant to call "function" a function which is not an analytic function defined by a single expression (vs piecewise functions). I'll try to find a reference. – Jean-Claude Arbaut Apr 23 '19 at 13:17
  • @Jean-ClaudeArbaut Ah okay thank you. – Ovi Apr 23 '19 at 13:54