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From the book, Advanced Engineering Mathematics, by Kreyszig,

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there quotes, "This differs completely from real calculus, Even if a real function is once differentiable we cannot conclude that it is twice differentiable nor that any of its higher derivatives exist."

What does this sentence mean? what could be particular example for this notion?

Does that mean in real calculus there could be a function that is differentiable once?

Or, does the real function may not have higher order differentiable function? what could be that function?

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    $\begingroup$ You can integrate any non-differentiable function, e.g., $y=|x|$. Then that function will be differentiable only once. $\endgroup$
    – velut luna
    Mar 1, 2016 at 15:03
  • $\begingroup$ See math.stackexchange.com/questions/78825/… By the way, in complex analysis, if a function is differentiable once, it's infinitely differentiable. $\endgroup$
    – Augustin
    Mar 1, 2016 at 15:04

4 Answers 4

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You can integrate any non-differentiable function and that function will be differentiable only once.

For example, the function

$$f(x)=\left\{ \begin{eqnarray} -x^2/2, & \ x < 0 \\ x^2/2, & \ x \ge 0 \end{eqnarray} \right.$$

is differentiable, but it's derivative

$$f'(x)=|x|$$ is not differentiable.

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Consider the function $f(x) = \left\{ \begin{eqnarray} -x^2, & \ x \le 0 \\ x^2, & \ x \gt 0 \end{eqnarray} \right.$. This is differentiable once, and its derivative is $f'(x) = \left\{ \begin{eqnarray} -2x, & \ x \le 0 \\ 2x, & \ x \gt 0 \end{eqnarray} \right. = 2 |x|$, which is clearly seen not to be differentiable in $0$ anymore. Such an annoying thing does not happen in complex analysis: if a function is once complex-differentiable, it will be complex-differentiable any number of times!

Even more: a function can be real-differentiable infinitely many times, yet fail to be real-analytic, the standard such example being $f(x) = \left\{ \begin{eqnarray} \textrm e ^{-\frac 1 {x^2}}, & \ x \le 0 \\ 0, & \ x \gt 0 \end{eqnarray} \right.$ Again, this does not happen in complex analysis: once a function is complex-differentiable, it is also complex-analytic! This shows that complex-differentiable functions are very "rigid", allowing their study to use plenty of algebraic techniques, opening the door towards some of the most beautiful branches of pure mathematics.

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There are continuous functions on $\mathbb R$ that are nowhere differentiable (Weierstrass). Let $f$ be such a function. Set $F(x)=\int_0^x f(t)\, dt.$ Then $F'(x)=f(x)$ everywhere, but $F''(x)$ exists nowhere.

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It means that there exist real valued functions that are $k$ times differentiable but not $k+1$ times differentiable for every $k$. An easy example of this is $|x|^k$ for odd $k$ and $|x|^{k-1}x$ for even $k$. Another way to do this is to integrate a non-differentiable function $k$ times

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    $\begingroup$ For odd $k$ otherwise you can omit $|.|$ $\endgroup$ Mar 1, 2016 at 15:09
  • $\begingroup$ @gammatester Yup, corrected $\endgroup$ Mar 1, 2016 at 15:11

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