The answer to this is probably quite obvious, but to be honest it hasn't clicked for me yet.
I can understand that for two curves f(x) and g(x) to be identical, f'(0)=g'(0), f''(0)=g''(0), f'''(0)=g'''(0) for all possible derivatives. But why is this a sufficient, as well as necessary, condition?
And when approximating, I understand why a straight line with m=f'(0) is better than y=f(0), but why is it that the best value for the x^2 coefficient is f''(0)/2!? Why does this result generalise?
Or is it not the best approximation? Would a Taylor approximation with a different centre give a better approximation?
As you can see, I'm a little confused, and it would be great if someone could help clear this up for me.

peterw55
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 16102017 11:10

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 16102017 12:10
(Original post by peterw55)
The answer to this is probably quite obvious, but to be honest it hasn't clicked for me yet.
I can understand that for two curves f(x) and g(x) to be identical, f'(0)=g'(0), f''(0)=g''(0), f'''(0)=g'''(0) for all possible derivatives. But why is this a sufficient, as well as necessary, condition?
And when approximating, I understand why a straight line with m=f'(0) is better than y=f(0), but why is it that the best value for the x^2 coefficient is f''(0)/2!? Why does this result generalise?
Or is it not the best approximation? Would a Taylor approximation with a different centre give a better approximation?
As you can see, I'm a little confused, and it would be great if someone could help clear this up for me.
However, I can (I hope) be a bit more illuminating with the coefficients. You wonder why, if the first two are f(0) and f'(0), the next one is f''(0)/2! rather than f''(0), and seem to be suggesting that this doesn't follow the pattern. But if you write the first two coefficients as f(0)/0! and f'(0)/1!, the pattern becomes clearer.
It's also not too hard to show that, if an infinite power series expansion is possible, these must be the coefficients. 
peterw55
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 16102017 12:14
(Original post by Pangol)
I can't give a good intuative explanation  the theory that I am familiar with assumes that an infinite power series expansion is possible, and then goes on to show why it must be found in the way that I am sure tha you know.
However, I can (I hope) be a bit more illuminating with the coefficients. You wonder why, if the first two are f(0) and f'(0), the next one is f''(0)/2! rather than f''(0), and seem to be suggesting that this doesn't follow the pattern. But if you write the first two coefficients as f(0)/0! and f'(0)/1!, the pattern becomes clearer.
It's also not too hard to show that, if an infinite power series expansion is possible, these must be the coefficients.
*I know it doesn't always, but for some curves the interval of validity is from infinity to infinity. 
uponthyhorse
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 16102017 12:17
Have you seen this video? It may not answer all your questions but this channel usually gives a good intuitive understanding of complex mathematical ideas.

peterw55
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 16102017 12:25
(Original post by uponthyhorse)
Have you seen this video? It may not answer all your questions but this channel usually gives a good intuitive understanding of complex mathematical ideas. 
DFranklin
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 16102017 12:26
(Original post by peterw55)
The answer to this is probably quite obvious, but to be honest it hasn't clicked for me yet.
I can understand that for two curves f(x) and g(x) to be identical, f'(0)=g'(0), f''(0)=g''(0), f'''(0)=g'''(0) for all possible derivatives. But why is this a sufficient, 
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 16102017 12:48
(Original post by Pangol)
I can't give a good intuative explanation  the theory that I am familiar with assumes that an infinite power series expansion is possible, and then goes on to show why it must be found in the way that I am sure tha you know.
However, I can (I hope) be a bit more illuminating with the coefficients. You wonder why, if the first two are f(0) and f'(0), the next one is f''(0)/2! rather than f''(0), and seem to be suggesting that this doesn't follow the pattern. But if you write the first two coefficients as f(0)/0! and f'(0)/1!, the pattern becomes clearer.
It's also not too hard to show that, if an infinite power series expansion is possible, these must be the coefficients.(Original post by peterw55)
Thanks for your answer. I do know why it's f''(0)/2! in the Maclaurin series (i.e. it follows the pattern). But I don't really know why an infinite series of increasing order derivatives can fully describe a curve*. If that is 'an infinite power series expansion', should I just take that as an assumption (i.e. it works graphically)?
*I know it doesn't always, but for some curves the interval of validity is from infinity to infinity.
You can then construct a polynomial P of degree at most n that satisfies p^(k)(a) = f^k(a) for k = 0,1,..., n1 and p(b) = f(b). (this is where the factorials come from).
Applying higher order Rolle's to this gives you f with the Lagrange form of the remainder which you can show goes to 0 as n > infinity for sufficiently nice functions f. 
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 16102017 12:51

DFranklin
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 16102017 12:53
(Original post by Zacken)
No, assuming that you can represent f by a power series is essentially dodging the question. What you need is some machinery: once you have your first order Rolle's theorem and you can assume that f is differentiable n times on an open interval containing [a,b] you get your higher order Rolle's theorem using induction and the first order Rolle's theorem. 
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 16102017 12:57
(Original post by DFranklin)
If you want something more accessible at Alevel, I think integrating n times by parts is a better bet. 
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 16102017 13:43
Proof of Maclauren expansion using integration by parts.
Assumption: f is n+1 times differentiable on [0,a].
, Here's the one "trick" (lecturers hate this!): we're going to integrate by parts with u = f'(x), dv = 1. But we're going to integrate dv to get xa, not x (remember we can add an arbitrary constant when we integrate). So
.
From this point we just integrate by parts normally:
etc until we get
(For a formal proof it's easier to start from this final expression and use induction, of course). 
peterw55
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 16102017 14:31
(Original post by DFranklin)
Proof of Maclauren expansion using integration by parts.
Assumption: f is n+1 times differentiable on [0,a].
, Here's the one "trick" (lecturers hate this!): we're going to integrate by parts with u = f'(x), dv = 1. But we're going to integrate dv to get xa, not x (remember we can add an arbitrary constant when we integrate). So
.
From this point we just integrate by parts normally:
etc until we get
(For a formal proof it's easier to start from this final expression and use induction, of course). 
DFranklin
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 16102017 14:53
(Original post by peterw55)
Thank you for typing all of this out. In the third line, how do you get that the integral of (xa) with respect to x is (xa)^2 / 2? 
peterw55
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 16102017 15:14
(Original post by DFranklin)
Standard result (for me). If you doubt it, differentiate (xa)^2 /2.
xa has antiderivative 0.5x^2  ax (ignoring the +c)
= x^2 / 2  ax
= x^2 / 2  2ax / 2
= x^2  2ax / 2
(xa)^2 / 1 = x^2 + a^2  2ax / 2
What stupid algebra mistake have I made? Where does the a^2 / 2 come from? 
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 16102017 15:16
(Original post by peterw55)
Yes, of course it works.
xa has antiderivative 0.5x^2  ax (ignoring the +c)
= x^2 / 2  ax
= x^2 / 2  2ax / 2
= x^2  2ax / 2
(xa)^2 / 1 = x^2 + a^2  2ax / 2
What stupid algebra mistake have I made? Where does the a^2 / 2 come from? 
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 16102017 15:19
(Original post by DFranklin)
Arbitrary constant. Both (x^2  2ax) /2 and (x^2  2ax +a^2) /2 have derivative (xa). 
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 16102017 15:22
(Original post by peterw55)
Because you can set a = 0?
Similarly, the integral of xa is not (x^22ax)/2, but (x^22ax/2) + C.
You have effectively taken C = 0, I have taken it to equal a^2/2. 
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 16102017 15:26
(Original post by DFranklin)
No. You should have had it drilled into you (to the point of exasperation!) that the integral of x is not x^2/2, but x^2/2+C. The point of this distinction (which is often not drilled nearly so well), is that any choice of C gives you a valid antiderivative.
Similarly, the integral of xa is not (x^22ax)/2, but (x^22ax/2) + C.
You have effectively taken C = 0, I have taken it to equal a^2/2. 
peterw55
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 16102017 15:35
(Original post by peterw55)
Of course! Now I deservingly feel stupid :/
Even if it's not intuitive, it's a good answer to my question, so thank you! 
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 16102017 15:46
(Original post by peterw55)
Even if it's not intuitive, it's a good answer to my question, so thank you!
Unfortunately, as the counterexample I posted shows, this turns out to not actually be the case. (In terms of the integral proof I showed you, the issue is that the final "integral" term doesn't get small enough to be ignored, no matter how many terms you take).
So at this point in your mathematical career, you're faced with something that you'd like intuitive justification for, and I can even given you a plausible intuitive argument, and yet it isn't actually true.
And then at university, when you study integration with complex numbers (not sure if you'll have done complex numbers at all yet), you find that if you have a complex function, and it's (complex) differentiable just once (for every point in a region), then you can make a Taylor expansion about any point in the region and all the derivatives will work and it will come out perfectly.
And of course at this point, your scepticism will have been tuned to the point where you really wouldn't expect this to work, and so your intuition gets fooled all over again...
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