## Help with a Tricky Derivative Proof

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Aeolus Gale
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Joined: Mon Nov 03, 2008 6:33 am UTC

### Help with a Tricky Derivative Proof

Hey all,
I've been having some trouble with the beginning of one of my homework problems for my real analysis class. The problem in question:

Suppose [imath]f : (a,b)\to\mathrm{R}[/imath] is differentiable at [imath]x\in(a,b)[/imath]. Prove that$\lim_{h\rightarrow 0} \frac{f(x + h) - f(x - h)} {2h}$exists and equals [imath]f'(x)[/imath].

Now, I tried messing around with the definition of the derivative of a function, but I couldn't really get anywhere. Could anyone point me in the right direction? Thanks :>

Stanford
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Joined: Fri Nov 02, 2007 6:52 pm UTC

### Re: Help with a Tricky Derivative Proof

Aeolus Gale wrote:Hey all,
I've been having some trouble with the beginning of one of my homework problems for my real analysis class. The problem in question:

Suppose [imath]f : (a,b)\to\mathrm{R}[/imath] is differentiable at [imath]x\in(a,b)[/imath]. Prove that$\lim_{h\rightarrow 0} \frac{f(x + h) - f(x - h)} {2h}$exists and equals [imath]f'(x)[/imath].

Now, I tried messing around with the definition of the derivative of a function, but I couldn't really get anywhere. Could anyone point me in the right direction? Thanks :>

You already know that $f'(x) = \lim_{h\rightarrow 0} \frac{f(x + h) - f(x)} {h}$ which is a similar form. Have you tried manipulating what you have to get exactly this equation, perhaps by adding and subtracting a specific quantity which it's missing, then working backwards?

Aeolus Gale
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Joined: Mon Nov 03, 2008 6:33 am UTC

### Re: Help with a Tricky Derivative Proof

I attempted to add and subtract f(x - h) from the numerator, and then I multiplied the whole thing by 2/2 to get the 2h on the bottom, but I couldn't figure out how to get it to look like that equation... I'm really at quite a loss here.

thornahawk
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### Re: Help with a Tricky Derivative Proof

BIG HINT:

First, you might want to show the truth of the so-called "backward difference form":

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

and then work from that and the other derivative formula.

~ Werner

P.S. Have you discussed Taylor's Theorem in the class? That's another route...
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Aeolus Gale
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Joined: Mon Nov 03, 2008 6:33 am UTC

### Re: Help with a Tricky Derivative Proof

thornahawk wrote:BIG HINT:

First, you might want to show the truth of the so-called "backward difference form":

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

and then work from that and the other derivative formula.

~ Werner

P.S. Have you discussed Taylor's Theorem in the class? That's another route...

How would I go about showing the truth of that form? I can certainly understand how that would help, but I don't think that I can use it if I can't prove it.

Charlie!
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Joined: Sat Jan 12, 2008 8:20 pm UTC

### Re: Help with a Tricky Derivative Proof

Aeolus Gale wrote:
thornahawk wrote:BIG HINT:

First, you might want to show the truth of the so-called "backward difference form":

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

and then work from that and the other derivative formula.

~ Werner

P.S. Have you discussed Taylor's Theorem in the class? That's another route...

How would I go about showing the truth of that form? I can certainly understand how that would help, but I don't think that I can use it if I can't prove it.

It's really just the the other definition of the derivative at x = (insert important bit here).

small hint: this form wouldn't work at a, because (a - h) doesn't necessarily behave well.

Note: unless, of course, I''m wrong and the easy way is invalid
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Yesila
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### Re: Help with a Tricky Derivative Proof

Aeolus Gale wrote:
thornahawk wrote:BIG HINT:

First, you might want to show the truth of the so-called "backward difference form":

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

and then work from that and the other derivative formula.

~ Werner

P.S. Have you discussed Taylor's Theorem in the class? That's another route...

How would I go about showing the truth of that form? I can certainly understand how that would help, but I don't think that I can use it if I can't prove it.

One way to prove that would be to look at one sided limits. and use the fact that h approaching 0 from the left is the same as -h approaching 0 from the right. Equate these one sided limits with the one sided limits in the "normal" derivative.

mike-l
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### Re: Help with a Tricky Derivative Proof

Charlie! wrote:
Aeolus Gale wrote:
thornahawk wrote:BIG HINT:

First, you might want to show the truth of the so-called "backward difference form":

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

and then work from that and the other derivative formula.

~ Werner

P.S. Have you discussed Taylor's Theorem in the class? That's another route...

How would I go about showing the truth of that form? I can certainly understand how that would help, but I don't think that I can use it if I can't prove it.

It's really just the the other definition of the derivative at x = (insert important bit here).

small hint: this form wouldn't work at a, because (a - h) doesn't necessarily behave well.

Note: unless, of course, I''m wrong and the easy way is invalid

This doesn't have any more or any fewer problems than a+h does at a, since h can be negative. Fortunately the original question was about the open interval (a,b) so we don't have to worry about end points at all.

To the OP, perhaps it would help to call h something else, say j, so the limit is (f(x) - f(x-j))/j, and then make a substitution to make that look more like (f(x+h)-f(x))/h.
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Hix
Posts: 364
Joined: Sun Oct 15, 2006 5:46 pm UTC

### Re: Help with a Tricky Derivative Proof

Aeolus Gale wrote:Hey all,
I've been having some trouble with the beginning of one of my homework problems for my real analysis class. The problem in question:

Suppose [imath]f : (a,b)\to\mathrm{R}[/imath] is differentiable at [imath]x\in(a,b)[/imath]. Prove that$\lim_{h\rightarrow 0} \frac{f(x + h) - f(x - h)} {2h}$exists and equals [imath]f'(x)[/imath].

Now, I tried messing around with the definition of the derivative of a function, but I couldn't really get anywhere. Could anyone point me in the right direction? Thanks :>

Since you are in a real analysis class, I guess that you may have encountered the definition of the derivative in terms of "little oh" notation? If you are comfortable with this notation, this problem is a rather fun application. To get you pointed in the right direction, start with arguing that since it is given that f is differentiable at x, you know that f(x+h) can be approximated as:

f(x+h) = f(x) + h*f'(x) + o(h)
[where o(h) represents some remainder function with the property that o(h)/h goes to zero as h goes to zero. This equation is basically the technical way of saying that points on the tangent line make a good approximation to points on the graph of f(x)]

If you know how to write down a similar expression that approximates f(x-h), then you'll find that taking the limit given in the homework problem becomes rather easy.

This is really exactly the same method that everyone else has been suggesting, but in different language. However, I know that some people find this notation easier to understand and manipulate than the limit definition of a derivative. So, if little oh notation has been covered in your class, look into using it!

Prediction: The next question in your homework (or perhaps a question on next week's homework) will ask you to prove that the derivative of any even function is an odd function, and that the derivative of any odd function is an even function.

Yakk
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### Re: Help with a Tricky Derivative Proof

Write out what "the derivatives exists, and is D" means, after you expand the definition of limit. (Ie, remove the word "limit" from the definition of the derivative, and demonstrate it is defining the same thing, by expanding what 'limit' means)

Write out what you need to show that the limit exists in the " / 2 h" case you are trying to prove. (Ie, remove the word "limit" from what you are trying to prove, and demonstrate that it is defining the same thing, by expanding what 'limit' means)

If you do not have an answer after doing that and thinking about it, post the result of doing that here.
One of the painful things about our time is that those who feel certainty are stupid, and those with any imagination and understanding are filled with doubt and indecision - BR

Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.