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Show $$\lim_{h \to \ 0} \frac{f(x + 2h) - 2f(x+h) + f(x)}{h^{2}} = f''(x)$$

Proof:

By definition:

$$f'(x) = \lim_{h \to \ 0} \frac{f(x + h) - f(x)}{h}$$

Using this idea it would imply:

$$1)\ \ f''(x) = \lim_{h \to \ 0} \frac{f'(x + h) - f'(x)}{h}$$

As such it is required that I find an expression for $f'(x+h)$. This is where I'm not sure if the step I took is legitimate.

An expression for $f'(x + h)$ is:

$$f'(x+h) = \lim_{h \to \ 0} \frac{f(x + 2h) - f(x + h)}{h}$$

Combining this with the definition of $f'(x)$ and inserting it into 1) you arrive at: $$\lim_{h \to \ 0} \frac{f(x + 2h) - 2f(x+h) + f(x)}{h^{2}} = f''(x)$$

As required.

Concern: I feel a discomfort with this solution. Even though "mechanically" it worked out, if I am taking the limit as $h \rightarrow 0$ that would mean $x + 2h$ and $x + h$ both go to $x$. But I am attempting to use the idea that $x +2h$ goes to $x + h$. Perhaps it is a notation idea that I need to communicate better, but I feel it is larger than just that.

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  • $\begingroup$ I feel the same way. Can't you use LH or Taylor? $\endgroup$
    – Botond
    Commented May 5, 2019 at 21:14
  • $\begingroup$ @Botond That would somewhat constrain the conditions under which the result is validated. $\endgroup$
    – J.G.
    Commented May 5, 2019 at 21:21
  • $\begingroup$ If you want to be really strict, then you should require your function $f$ to be differentiable on a neighborhood of $x$, not just differentiable at $x$. $\endgroup$
    – Firepi
    Commented May 5, 2019 at 21:23
  • $\begingroup$ @J.G. Why would it? I think the existence of the second derivative for the limit in the title is enough, and Taylor's theorem with Peano's remainder should work in that case. $\endgroup$
    – Botond
    Commented May 5, 2019 at 21:25
  • $\begingroup$ I think the MVT might work, but it'd require $2$ times differentiability around $x$. $\endgroup$
    – Botond
    Commented May 5, 2019 at 22:37

3 Answers 3

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We have $$f^{\prime}(x+h)-f^{\prime}(x)=\lim_{k\to0}\frac{f(x+h+k)-f(x+h)}{k}-\lim_{k\to0}\frac{f(x+k)-f(x)}{k}\\=\lim_{k\to0}\frac{f(x+h+k)-f(x+h)-f(x+k)+f(x)}{k}$$and so$$f^{\prime\prime}(x)=\lim_{h\to0}\lim_{k\to0}\frac{f(x+h+k)-f(x+h)-f(x+k)+f(x)}{kh}.$$You're right to feel "discomfort": we need to justify how we go from using two distinct tend-to-$0$ variables to just one. Here's my understanding (but a greater expert on analysis might say this isn't the right way to do it):

The quantity whose double limit is taken is $h\leftrightarrow k$-symmetric, so iff $f^{\prime\prime}(x)$ exists we can unambiguously write$$f^{\prime\prime}(x)=\lim_{h,\,k\to0}\frac{f(x+h+k)-f(x+h)-f(x+k)+f(x)}{kh}.$$Then we can take $h,\,k$ to $0$ in any way we like, all giving the same result, so let's take $k=h$. Then$$f^{\prime\prime}(x)=\lim_{h\to0}\frac{f(x+2h)-2f(x+h)+f(x)}{h^2}.$$

Edit: taking advice from comments @ParamanandSingh left under the OP and this answer, let's do it in a less suspect way, which does use just one tending-to-$0$ variable but not in the way the OP tried. Using L'Hôpital's rule,$$\lim_{h\to 0}\frac{f(x+2h)-2f(x+h)+f(x)}{h^2}=\lim_{h\to 0}\frac{f^\prime(x+2h)-f^\prime(x+h)}{h}\\=2\lim_{h\to 0}\frac{f^\prime(x+2h)-f^\prime(x)}{2h}-\lim_{h\to 0}\frac{f^\prime(x+h)-f^\prime(x)}{h}=2f^{\prime\prime}(x)-f^{\prime\prime}(x)=f^{\prime\prime}(x).$$

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  • $\begingroup$ Thanks for articulating what I couldn't put into words. The need to justify those two tend to 0 variables is what has me perturbed. I feel as though you do have the right idea though. $\endgroup$ Commented May 5, 2019 at 21:31
  • $\begingroup$ @ParamanandSingh and J.G, how is this valid if I'm supposed to be assuming I can only use the "fundamental" definition of the derivative? In such a scenario I don't have L'Hopital's Rule at my disposal...... $\endgroup$ Commented May 12, 2019 at 21:39
  • $\begingroup$ @dc3rd: you can't do it just via definition of derivative. You need to use mean value theorem or L'Hospital's Rule or Taylor series. Moreover you did not mention to avoid L'Hospital's Rule in your question. $\endgroup$
    – Paramanand Singh
    Commented May 13, 2019 at 2:17
  • $\begingroup$ @ParamanandSingh The source I got the question from did not mention we could not use any of these techniques, I just made the assumption base doff of building from the fundamental pieces that there would be a way to derive the result. It is something that I get in trouble with a lot when attempting to construct solutions....GOing to have to learn when to pull back and use the refined machinery available to me. Thanks for the help. $\endgroup$ Commented May 13, 2019 at 16:44
  • $\begingroup$ @dc3rd: it is always better to avoid advanced machinery if possible (Jacobi and Ramanujan are great examples who used simple machinery to obtained very deep and difficult results). Unfortunately such an attitude has now become unfashionable. $\endgroup$
    – Paramanand Singh
    Commented May 13, 2019 at 17:45
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Let $f$ be any continuous function which is second differentiable at $a$ (i.e. $f''(a)$ exists). By definition of second differentiability at a point, $f'(x)$ exists over $(a-\eta,a+\eta)$ for some $\eta > 0$. Define $g : (-\eta, \eta) \to \mathbb{R}$ by

$$g(x) = f(a+x) - f(a) - f'(a) x - \frac12 f''(a) x^2$$

It is clear $g$ is continuous, second differentiable at $a$ and $g'(x)$ exists over $(-\eta,\eta)$. Furthermore, $g(0) = g'(0) = g''(0) = 0$. What we want to show is equivalent to following statement

$$\lim_{h\to 0}\frac{g(2h) - 2g(h) + g(0)}{h^2} = 0 \tag{*1}$$

Since $f''(a)$ exists, $g''(0) = \lim_{h\to 0}\frac{g'(h)}{h}$ exists and vanishes. For any $\epsilon > 0$, there exists a $\delta \in ( 0,\eta)$ such that

$$\left|\frac{g'(x)}{x}\right| = \left|\frac{g'(x) - g'(0)}{x}\right| < \frac{\epsilon}{3}\quad\text{ whenever }\quad 0 < |x| < \delta$$

For any $h$ such that $0 < |h| < \frac{\delta}{2}$, apply MVT to $g(2h) - g(h)$ and $g(h) - g(0)$, we find there are $p,q \in (0,1)$ such that $$\frac{g(2h) - g(h)}{h} = g'((1+p)h)\quad\text{ and }\quad \frac{g(h) - g(0)}{h} = g'(qh)$$

This leads to

$$\begin{align}\left|\frac{g(2h) - 2g(h) + g(0)}{h^2}\right| & \le \left| \frac{g(2h) - g(h)}{h^2}\right| + \left|\frac{g(h) - g(0)}{h^2}\right|\\ &= \left|\frac{g'((1+p)h)}{h}\right| + \left|\frac{g'(qh)}{h}\right| \\ & < \frac{\epsilon}{3}\left((1+p) + q\right)\\ & < \epsilon \end{align} $$

Since $\epsilon$ can be arbitrary small, $(*1)$ follows and hence $$\lim_{h\to 0}\frac{f(a+2h) - 2f(a+h) + f(a)}{h^2} = f''(a)$$

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  • $\begingroup$ Nice proof going back to basics. +1 $\endgroup$
    – Paramanand Singh
    Commented May 6, 2019 at 9:38
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Yes, that expression is correct. You are finding the derivative at the point $y = x+h$ usinh the old definition

$$ f'(y) = \lim_{ h \to 0} \frac{ f(y + h) - f(y) }{h} $$

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    $\begingroup$ Ok. But shouldn't the two h's be distinguished ? Because what I gather you are saying is that: $x + 2h = (x + h) + h$. In which case I am treating $(x + h)$ as my fixed point while approaching from the distance "h". That seems like it could get confusing. $\endgroup$ Commented May 5, 2019 at 20:45

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