# I Limits to directly check second order differentiability

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1. Feb 28, 2017

### Kumar8434

Can the equality of these two be used as a second-order differentiability check?:

$$\lim_{h\rightarrow 0}\frac{f(x+2h)-2f(x+h)+f(x)}{h^2}$$
And,
$$\lim_{h\rightarrow 0}\frac{f(x-2h)-2f(x-h)+f(x)}{h^2}$$

I think that just like when the limits $\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$ and $\lim_{h\rightarrow 0}\frac{f(x-h)-f(x)}{-h}$ are equal, then it means that the function is differentiable, the equality of these two limits should mean that the function is twice differentiable.

Last edited: Feb 28, 2017
2. Mar 2, 2017

### Stephen Tashi

Those two limits are the same limit.

Those two limits are also the same limit.

I think you mean to distinguish between limits like $lim_{h \rightarrow 0^+} g(h)$ and $\lim_{h \rightarrow 0^{-}} g(h)$ , but you can't do this by a limit of the form $lim_{h \rightarrow 0} g(h)$ because there is nothing in the definition of that limit that distinguishes a "side" to the limit. The variable $h$ does not necessarily denote a positive quantity, so $x + h$ is not necessarily to the right of $x$.

A definition of $lim_{h \rightarrow 0} g(h) = L$ is:
For each real number $\epsilon > 0$ there exits a number $\delta> 0$ such that if $h \in (0-\delta, 0 + \delta)$ and $h \ne 0$ then $g(x) \in ( L - \epsilon, L + \epsilon)$.

In that definition $h$ can be to the right or left of zero. It is only specified that $h$ is in an interval about zero and not equal to zero.

By contrast, the definition of $\lim_{h \rightarrow 0^+} g(h) = L$ can be stated as:
For each real number $\epsilon > 0$ there exits a number $\delta> 0$ such that if $h \in (0, 0 + \delta)$ then $g(x) \in ( L - \epsilon, L + \epsilon)$.

3. Mar 2, 2017

### BvU

I think there's also a notation that uses $\uparrow$ and $\downarrow$ in e.g. $$\displaystyle {\lim_{h\downarrow 0}} ...$$

4. Mar 3, 2017

### Kumar8434

I meant $h$ is already assumed to be positive, so I guess I meant to write:
$$\lim_{h\rightarrow 0^+}\frac{f(x+2h)-2f(x+h)+f(x)}{h^2}$$
And,
$$\lim_{h\rightarrow 0^+}\frac{f(x-2h)-2f(x-h)+f(x)}{h^2}$$
The first limit gives you the Right hand second order derivative and the second limit gives you the Left hand second order derivative.
I checked it for this function:
$f(x)=\frac{x^2}{2}, x\geq0;$
$=\frac{-x^2}{2}, x<0;$
This function is not twice differentiable at $x=0$ and the limits confirmed that.
I also just got this general rule by a similar method for a function to be $n$ times differentiable:
$$\lim_{h\rightarrow 0^+}\frac{\sum_{r=0}^{n}(-1)^r\cdot \binom{n}{r}\cdot f(x-(n-r)h)}{(-h)^n}=\lim_{h\rightarrow 0^+}\frac{\sum_{r=0}^n(-1)^r\cdot \binom{n}{r}\cdot f(x+(n-r)h)}{h^n}$$
$h$ itself is assumed to be positive, that's why I've written $h\rightarrow 0^+$. The function should be first $n-1$ times differentiable for this rule to check $n$ times differentiability. Even if the function is not $n$ times differentiable, then the LHS of this expression gives the $n^{th}$ Left-hand derivative and the RHS gives $n^{th}$ Right hand derivative.

Last edited: Mar 3, 2017
5. Mar 3, 2017

### Stephen Tashi

Let's formulate a specific mathematical question:

Assume
$$\lim_{h\rightarrow 0^+}\frac{f(x+2h)-2f(x+h)+f(x)}{h^2}$$ exists and is equal to $L_1$.
$$\lim_{h\rightarrow 0^+}\frac{f(x-2h)-2f(x-h)+f(x)}{h^2}$$ exists and is equal to $L_2$.
Assume $L_1 = L_2$.

Then is it true that $f''(x)$ exists (at the value $x$) and that $f''(x) = L_1 = L_2$ ?
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Before trying to settle that question, we can say: Yes, you have a correct intuitive idea. In the study of numerical analysis, there are "multi point methods" for approximating derivatives, including higher derivatives. The methods that are easier to understand use evenly spaced points ( such as x , x + h, x + 2h). ( https://en.wikipedia.org/wiki/Finite_difference_coefficient , https://en.wikipedia.org/wiki/Numerical_differentiation ). There are also methods that can used arbitrarily spaced points (http://www3.nd.edu/~zxu2/acms40390F11/sec4-1-derivative-approximation-1.pdf ).

Unfortunately, for the purpose of answering the above theoretical question, most explanations of these methods simply assume the existence of the derivative to be approximated.

It's tempting to answer the theoretical question by apply some of the handy "rules of limits" such as $lim_{x \rightarrow a} ( f(x) + g(x)) = lim_{x \rightarrow a} f(x) + lim_{x \rightarrow a} g(x)$. However, such that rule only applies if the two limits on the right hand side of the equation are assumed to exist. (e.g. Let $f(x) = 1$ if $x$ is a rational number and $f(x) = 0$ otherwise. let $g(x) = 1- f(x)$ Then $lim_{x \rightarrow 0} (f(x) + g(x)) = 1$, but neither $lim_{x \rightarrow 0} f(x)$ nor $lim_{x \rightarrow 0} g(x)$ exist.

So I don't know the answer to the above question "off the top of my head". I'll have to think about.

The kind of situation to worry about is a function that has a first derivative, but no second derivative - such as https://calculus.subwiki.org/wiki/Derivative_of_differentiable_function_need_not_be_continuous

One simplification to the question is to note that:
$$\lim_{h\rightarrow 0^+}\frac{f(x-2h)-2f(x-h)+f(x)}{h^2} = lim_{h \rightarrow 0^-} \frac{f(x+2h)-2f(x+h)+f(x)}{h^2}$$, which can be show by using the definitions of the limits involved. So the assumptions of the question can be simplified to :

Assume $lim_{h \rightarrow 0} \frac{f(x+2h)-2f(x+h)+f(x)}{h^2}$ exists.

since that assumption is equivalent to assuming the two "one sided" limits exist and are equal.

6. Mar 3, 2017

### Kumar8434

I think it definitely works for functions whose first derivative exists but second derivative doesn't exist. ;But, if we look at the function $|x|$, then it is not even once differentiable, so it can't be twice differentiable, but both these limits evaluate are equal to 0. That's why I've written that, for this rule to check $n^{th}$ differentiability, the function should be first $n-1$ times differentiable. But still it's worth noting that the left hand and right hand derivatives of |x| are -1, and 1. Differentiating these both separately gives 0 and 0, which is what the limits gave.

7. Mar 3, 2017

### Stephen Tashi

Can you explain what you mean by "it"? What precisely are you claiming is true?

8. Mar 5, 2017

### Kumar8434

I'm claiming that the limits are always unequal when the first derivative of a function exists but the second derivative does not.

9. Mar 5, 2017

### Stephen Tashi

Ok, let's make that a precise claim about a real valued function $f(x)$ of a real variable:

Claim: if $f'(x)$ exists at $x = a$ and $f''(x)$ does not exist at $x = a$ then:
$lim_{x \rightarrow h^+} \frac{ f(a+2h) -2f(a+h) + f(a)}{h^2}$ exists and
$lim_{x \rightarrow h^-}\frac{f(a+2h)-2f(a+h) + f(a)}{h^2}$ exists
and the two limits are unequal.

Is that the claim? Or do we wish not to claim that the two limits must exist ?

10. Mar 5, 2017

### Kumar8434

No, this is the claim:
If $f'(x)$ exists at $x=a$, then $f''(x)$ exists at $x=a$ only if these two limits exist and are equal:
$$\lim_{h\rightarrow 0^+}\frac{f(a+2h)-2f(a+h)+f(a)}{h^2}$$
And
$$\lim_{h\rightarrow 0^+}\frac{f(a+2h)-2f(a+h)+f(a)}{h^2}$$

11. Mar 6, 2017

### Stephen Tashi

By the usual interpretation of the pattern "statement_B only if statement_C", it is equivalent to "if statement_B then statement_C".

The general pattern of the claim is : If statement_A then (statement_B only if statement C). This is logically equivalent to "If statement_A then (if statement_B then statement_C)", which is equivalent to " If (statement_A and statement_B) then statement_C". So the claim is equivalent to:

If $f'(x)$ and $f"(x)$ exist at $x=a$ then these two limits exist and are equal:
$$\lim_{h\rightarrow 0^+}\frac{f(a+2h)-2f(a+h)+f(a)}{h^2}$$
and
$$\lim_{h\rightarrow 0^-}\frac{f(a+2h)-2f(a+h)+f(a)}{h^2}$$