Interesting differentiation question
If, say, we have two functions, h and g, and h(x)≈g(x), would it follow that h'(x)≈g'(x) if we took d/dx of both sides, or does it apply only when there is strict equality of the two functions?
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Original post by DeuteriumPie
If, say, we have two functions, h and g, and h(x)≈g(x), would it follow that h'(x)≈g'(x) if we took d/dx of both sides, or does it apply only when there is strict equality of the two functions?
What do you mean by h(x)≈g(x) ?
Original post by DeuteriumPie
If, say, we have two functions, h and g, and h(x)≈g(x), would it follow that h'(x)≈g'(x) if we took d/dx of both sides, or does it apply only when there is strict equality of the two functions?
I wouldn't have thought so, depending on how strictly you define ≈. You could have two functions that are almost identical in terms of values but have a lot of small-scale (fractal-like) detail, resulting in totally different gradients.
Original post by Plagioclase
I wouldn't have thought so, depending on how strictly you define ≈. You could have two functions that are almost identical in terms of values but have a lot of small-scale (fractal-like) detail, resulting in totally different gradients.
In contrast, I think with integration you can use that approximation. (Can someone verify?)
Original post by Alex:
In contrast, I think with integration you can use that approximation. (Can someone verify?)
That makes intuitive sense to me, although I have no idea how you'd verify it.
Original post by notnek
What do you mean by h(x)≈g(x) ?
Okay, not entirely clear since u could take it all onto once side and make a new function out of it lol
Ok, say we had some expression that is APPROXIMATELY EQUAL to some other. Could we take d/dx of both sides, and maintain the approximate equality? I suspect no
Original post by Plagioclase
That makes intuitive sense to me, although I have no idea how you'd verify it.
A bit wishy washy, but the reason why the trapezium rule and simpson's rule exist is because integration is well behaved and easy to approximate; whereas with differentiation, no real alternative exists.
Image - choosing smaller trapeziums in the trapezium rule always gives you a more accurate answer, but this image indicates that this is not the case with numerical differentiation.
(edited 8 years ago)
Original post by Plagioclase
I wouldn't have thought so, depending on how strictly you define ≈. You could have two functions that are almost identical in terms of values but have a lot of small-scale (fractal-like) detail, resulting in totally different gradients.
Yeah, my exact thoughts!
From this page: http://mathworld.wolfram.com/NumericalDifferentiation.html
Pretty cool stuff if you want to look deeper into it.
Numerical differentiation is the process of finding the numerical value of a derivative of a given function at a given point. In general, numerical differentiation is more difficult than numerical integration. This is because while numerical integration requires only good continuity properties of the function being integrated, numerical differentiation requires more complicated properties such as Lipschitz classes.
Pretty cool stuff if you want to look deeper into it.
Original post by kkboyk
What about h''(x)≈g''(x)?
I'd imagine that's even more inaccurate, since you're differentiating twice!
Original post by Alex:
I'd imagine that's even more inaccurate, since you're differentiating twice!
Oh yeah, you're right I didn't see that graph you've posted
Immediately, my thoughts are yes, it could be.
Differentiability is a strong form of continuity in a sense. If f and g are "close" always, then their derivatives can't differ too much on some scale as otherwise one would of course grow too much faster than the other or something. It depends how close is close!
edit: nope, thinking of l'hoptials was downright stupid.
Differentiability is a strong form of continuity in a sense. If f and g are "close" always, then their derivatives can't differ too much on some scale as otherwise one would of course grow too much faster than the other or something. It depends how close is close!
edit: nope, thinking of l'hoptials was downright stupid.
(edited 8 years ago)
Weird, i thought it wouls be. Can someone give a strong example of why not?
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i dont think the diffrential would be equivalent as if you think of two curves h(x) and g(x) intersecting at one point, they will both be equal each other but if you look at their gradients h'(x) and g'(x) one may have a negative gradient while the other has a positive. idno just a thought lol
Original post by dijam
i dont think the diffrential would be equivalent as if you think of two curves h(x) and g(x) intersecting at one point, they will both be equal each other but if you look at their gradients h'(x) and g'(x) one may have a negative gradient while the other has a positive. idno just a thought lol
They are not just at one point. I assume it means for all x in the domain.
What your saying would just be g(a) aprrox h(a) for some a. That a cant take all values in the domain and just approx for 1 point.
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(edited 8 years ago)
Original post by physicsmaths
They are not just at one point. I assume it means for all x in the domain.
Thsi would just be g(a) aprrox h(a) for some a.
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Thsi would just be g(a) aprrox h(a) for some a.
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ooops was thinking of it wrong! silly me
Original post by DeuteriumPie
If, say, we have two functions, h and g, and h(x)≈g(x), would it follow that h'(x)≈g'(x) if we took d/dx of both sides, or does it apply only when there is strict equality of the two functions?
We can say h(x)=g(x)+O
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Original post by physicsmaths
Damn, this is weird. I had a similar idea brushing my teeth just now haha
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