For tiny h, f(x+h) = f(x) + hf'(x) ?

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For tiny h, f(x+h) = f(x) + hf'(x) ??

Hi all
I've been reading about proof of the chain rule and something is making me not sleep at night..
How is that possible that: "for tiny h, f(x+h) = f(x) + hf'(x)" ?


Even if 'h' is ultra-small, then "f(x+h)" will always differ from "f(x) + hf'(x)"... I know - the smaller the 'h', the smaller the difference but the difference will always exist for 'h' not equal to zero... So how can we plug this: "f(x) + hf'(x)" instead of "f(x+h)"..
Can someone explain this to me?
 

Answers and Replies

  • #2
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It comes directly from the definition of the derivative. Write f'(x) as follows:

[tex]f'(x) = \lim_{h \rightarrow 0} \frac{f(x + h) - f(x)}{h}[/tex]

The limit is taken as h goes to 0, i.e. |h| is small but not zero. We can rewrite the limit as an approximation for small h:

[tex]f'(x) \approx \frac{f(x + h) - f(x)}{h}[/tex]
[tex]\implies hf'(x) \approx f(x + h) - f(x)[/tex]
[tex]\implies hf'(x) + f(x) \approx f(x + h).[/tex]

When h gets smaller and smaller, the [itex]\approx[/itex] becomes [itex]=[/itex].
 
  • #3
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I don't like this. Not even for tiny h does the equality necessarily hold. The correct way of saying this is "for tiny h, [itex]f(x+h)\sim f(x)+hf^\prime(x)[/itex]. Writing an equality there is just wrong...
 
  • #4
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When h gets smaller and smaller, the [itex]\approx[/itex] becomes [itex]=[/itex].

But how it can become '=' ? I know that the error will become smaller and smaller as 'h' goes to zero, but my way of thinking is that we can never replace '[itex]\approx[/itex]' with '[itex]=[/itex]'... Even if 'h' = 0,000000000000(...)0001 the error will always exist. I just want to be precise.
 
  • #5
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But how it can become '=' ? I know that the error will become smaller and smaller as 'h' goes to zero, but my way of thinking is that we can never replace '[itex]\approx[/itex]' with '[itex]=[/itex]'... Even if 'h' = 0,000000000000(...)0001 the error will always exist. I just want to be precise.

You are correct. The equality will never hold. There always is an error. That's why writing f(x+h)=f(x)+hf'(x) is wrong.
 
  • #6
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By "smaller and smaller", I mean "in the limit as h approaches zero". In a lot of applications, using "tiny values" is more useful even though there is loss of precision. Of course there is always an error in difference. that's the definition of limit.
 
  • #7
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You are correct. The equality will never hold. There always is an error. That's why writing f(x+h)=f(x)+hf'(x) is wrong.

They use this equation in so many texts... It seems that these texts are massively erroneous...
 
  • #8
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^ Not massively. Only by a small error. ;)

The appropriate amount of rigor depends on context. Engineers will use this "equation" because it helps in applications. Opticians frequently use [itex]\theta[/itex] in place of [itex]\sin\theta[/itex] when [itex]|\theta|[/itex] is small. It's less precise for sure, but it's a useful physical approximation.
 
  • #9
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They use this equation in so many texts... It seems that these texts are massively erroneous...

From a mathematical point-of-view, those texts are indeed wrong. But the error is so small and insignificant, that people like to use the equation because it gives good approximations.

But, as a mathematician, I would never write something like this...
 
  • #10
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^ You say that like one cannot call themselves a mathematician if they write that. I'd write it if someone asks why it is written, which is exactly what's going on in this thread.

I wouldn't say it's wrong. Just not 100% mathematically precise. There is room for error (literally) in the context that something like this is used.
 
  • #11
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^ You say that like one cannot call themselves a mathematician if they write that. I'd write it if someone asks why it is written, which is exactly what's going on in this thread.

I said that "I, as a mathematician, would never write it like that". You can do what you want to however.

I wouldn't say it's wrong. Just not 100% mathematically precise. There is room for error (literally) in the context that something like this is used.

If it's not 100% mathematically precise, then it's wrong. Writing an equality there is wrong, in my opinion. It's ok to write [itex]\sim[/itex] or to do something approximate, but an equality is wrong.

Don't misunderstand me, I know many texts write an equality. Why? Because it works. In practice, the error is very small so the equality won't matter. But in theory the equality is wrong. And I would never write that.
 
  • #12
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I see that you were referring to yourself there. But you make your point by saying that you are a mathematician, justifying that you should not write that. It's pretty clear that you're just saying "a mathematician should not write something like this". Of course I don't disagree with you in that equality should not be written.

Here, "[itex]=[/itex]" doesn't really mean the two things are the indistinguishable, without error. It should be taken to mean that one expression can be replaced by another in a particular situation. That particular situation is specified -- it's when h is small.

I must have misinterpreted the OP post as asking for an explanation of this equation (or rather, approximation), when he was actually asking about why equality should hold.
 
  • #13
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In my opinion , writing an '=' sign is pretty much "massively erroneous". The whole point is that you can approximate the difference with L ( h ) , not that it necessarily actually becomes L ( h ). In my mind, it undermines the whole idea behind "limits" and "approximating", so it's a pretty bad thing to write.
 
  • #14
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It's pretty inaccurate in a rigorous mathematical context, for sure. But I bet the book you were reading wasn't really oriented for pure math, rather for engineering or science, amirite?
 
  • #15
disregardthat
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Using the approximation symbol without reference to how approximate it is is ambiguous and any mathematician would (definitely) not do this if his motive was not to e.g. make it easier to understand for people not entirely comfortable with the theory behind it.
 
  • #16
lurflurf
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we have, f(x+h) = f(x) + hf'(x+a ) for some a in (-|h|,|h|)
Often if h is tiny we can make use of the equation for some purpose without knowing a.
The fact that a is almost zero and that its specific value is not important does not mean a=0, usually it does not.
 
Last edited:
  • #17
HallsofIvy
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If you are working in "non-standard" analysis, where we have "infinitesmals" and say, not "tiny h", but "infinitesmal h", then it is, in fact, true that f(x+ h)= f(x)+ hf'(x)
 
  • #18
disregardthat
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If you are working in "non-standard" analysis, where we have "infinitesmals" and say, not "tiny h", but "infinitesmal h", then it is, in fact, true that f(x+ h)= f(x)+ hf'(x)

You are forgetting to take the standard part here.
 
  • #19
chiro
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I am with micromass on this one, wrong is wrong, and it would be misleading to say that the equality is actually an equality.

The best way to communicate this is to add the order term to the definition, to denote the error term.
 
  • #20
lurflurf
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^
x+x=2x
is wrong anyone can see there is no plus on then right hand side.
An equal sign means the two sides are equivalent in some sense.
This can be confusing at times, but it is also confusing to require the equivelance to be explicit.
mathematics: A bad habit of considering things that are basically the same to be different, whilst simultaneously considering things that are basically different to be the same.
 
  • #21
pwsnafu
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^
x+x=2x
is wrong anyone can see there is no plus on then right hand side.
An equal sign means the two sides are equivalent in some sense.
This can be confusing at times, but it is also confusing to require the equivelance to be explicit.
mathematics: A bad habit of considering things that are basically the same to be different, whilst simultaneously considering things that are basically different to be the same.

I honestly don't know if this is a troll, but considering the post count I don't think so.

In mathematics, one of the most important things we do is define what equality mean in a space. In a polynomial ring, we define the symbol 2x to be x+x. So therefore the equality is valid. In the space of real functions, equality means that the values evaluated are equal and the domain and codomain are equal. This is not satisfied in OP's equation.

You can make snarky comments all you want, but mathematics is about rigor. You can't change that.
 

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