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Okai, so we were learning about Newtons method of differentiation and I came to questions why Isaac Newton or Leibniz use the following function.

$${f(x+h) - f(x)\over h}$$

$h$ is the distance at the X axis of the point we wish to find. However, this equation requires us to use a function $\lim$ which really means that the number $h$ is of such small significance that we simply remove it from the equation. Which in my eyes is more about relativity than maths. Lets say we use the function = $$x^2+4x$$ We all know that derivative is = $$2x + 4$$ So I came up with a solution like this.

$${f(x+h) - f(x-h)\over 2h}$$

Why dont we use this method instead of learning to use a method that requires us to use relativity and take it into consideration?

EDIT:

$$f(x+h) = (x+h)^2 + 4(x+h)$$ $$f(x-h) = (x-h)^2 + 4(x-h)$$

Derivate is:

$$(((x+h)^2 + 4(x+h)) - ((x-h)^2 + 4(x-h))/h)\over 2$$

which gives out to be = 2x + 4 once all the calculation has been carried out.

Cameron Buie
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Gjert
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2 Answers2

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Your method gives us what is known as the symmetric derivative (at least, if you take it to the limit). If you'll read over the article, you'll see that such a beast may be defined where $f$ isn't, or where $f$ is defined but not continuous, or where $f$ is defined and continuous but has a "sharp corner". This is probably why differentiability is preferred in many circumstances (symmetric differentiability is "too weak"), even though the two types of derivatives agree at points of differentiability.

If you don't take the limit with your expression, then all you've given is the slope of a particular sort of secant line. This tells us little in general (if anything) about the behavior of the function at the point in question, unless we take $h$ to be small enough (that is, take the limit as $h\to 0$). For example, consider $f(x)=x^3$. A perfectly friendly function, no? But $$\frac{f(x+h)-f(x-h)}{2h}=3x^2+h^2,$$ which is not the same as the derivative (and may be very different!), unless you take the limit as $h\to 0$.

Cameron Buie
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  • It does give the slope of the tangent line in the limit that $h$ goes to $0$ (provided the function is nice). What FriedBitz wrote down is a central difference instead of right (left) difference. You can see that it gives the slope of the tangent line (provided it exists) because you can add and subtract $f(x)$ to get $\frac{1}{2}\left(\frac{f(x+h)-f(x)}{h}+\frac{f(x)-f(x-h)}{h}\right)$. In the limit that $h$ goes to $0$, we have $f'(x)$. – Cameron L. Williams Jun 17 '13 at 02:18
  • @CameronWilliams: In the post I saw originally, Jack M had fixed the OP's format, and in so doing, had replaced the $x-h$ with $x$. Answer now fixed to answer OP's original question. – Cameron Buie Jun 17 '13 at 02:23
  • However, wouldn't this tell us exacly the same of the point as a non symmetric derivation would? We do end up with the same result. – Gjert Jun 17 '13 at 02:31
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    @FriedBitz: Not in general. I suspect that only works for polynomials, and then, only those of degree at most $2$. See my updated answer. – Cameron Buie Jun 17 '13 at 02:32
  • @FriedBitz: I have confirmed that it works for polynomials of degree at most $2$, and that it fails for all other polynomials. I'm not going to bother to prove that it fails for non-polynomial functions. I will say kudos to you for noticing that this works in your example (and I have upvoted your question)! – Cameron Buie Jun 17 '13 at 02:46
  • Okai, I shall see if it works further on :) Thanks for good description. Just confused me a bit why our teacher just said it was impossible without even looking or listening to me... as if finding another method for the exact problem was impossible. but thanks :) – Gjert Jun 17 '13 at 02:50
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Your proposed derivative $f'$ of a function $f$ is defined by:

$$f'(x)={f(x+h)-f(x-h) \over 2h}$$

But you never specify what $h$ should be equal to, so in effect this is not a definition. You have basically defined an infinite number of derivatives, depending on the value of $h$ used. Nevertheless, let's take a look at your example calculation.

I assume that here you're basically considering the function $f(x)=x^2+4x$. Applying your definition, we get a derivative of ${x^2 \over h}+2x+2$ - unless I'm more tired than I thought, the result you edited into your question is wrong. So even in the derivative, we find $h$, and since you haven't specified what $h$ should be equal to, that means the derivative isn't really defined. It could be anything, depending on what value you choose for $h$.

Of course, the real reason why we use the limit definition is because there is a real world concept that we are trying to capture with the notion of a derivative. The limit definition accurately represents that concept, your proposed definition does not. How did you come up with, to begin with?

Jack M
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  • It appears that the OP discovered in the example $f(x)=x^2+4x$ that $$f'(x)=\frac{f(x+h)-f(x-h)}{2h}$$ for all $h\ne0$ (this holds in general for polynomials of degree at most $2$), which prompted the question. Of course, it doesn't work for other functions. – Cameron Buie Jun 17 '13 at 02:44
  • the H gets taken out, which leaves the example left with 2x+4... I did the caluclation on paper... exact same one. – Gjert Jun 17 '13 at 02:45
  • @FriedBitz but it doesn't work from $\cos(x)$, $\log(x)$, or $x^3$ – Ink Jun 17 '13 at 03:21
  • Yeah, Cameron told me and I tried myself, however, what does the leftover h mean if I tried by using x^3? – Gjert Jun 17 '13 at 04:17
  • @FriedBitz: It means that the slope of the secant line between the points $\langle x-h,(x-h)^3\rangle$ and $\langle x+h,(x+h)^3\rangle$ depends not only on the "center" of your interval ($x$), but also on its "radius" ($h$). For polynomials of degree less than $3$, this is not the case--all that matters there is the "center". – Cameron Buie Jul 04 '13 at 00:11