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Intuitively, why does the Maclaurin expansion work? Watch

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    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.
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    (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.
    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.
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    (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.
    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.
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    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.
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    (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.
    Thanks for the link. I've seen some videos from that channel before, but not this one. I'll watch it after I finish the C4 paper I'm doing :-)
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    (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,
    It is not sufficient. It is possible for 2 functions to agree for all derivatives when x = 0 and yet be different everywhere else. The standard counterexample is \displaystyle f(x) = \left\{ \begin{array}{@{}l@{\thinspace}l  }\exp(-1/x^2) &\quad x\neq 0 \\ 0 & \quad x = 0\end{array}\right which is infinitely differentiable everywhere, f^{(n)}(0) = 0 for all n, but f is clearly non-zero everywhere except x=0.
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    (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.
    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.

    You can then construct a polynomial P of degree at most n that satisfies p^(k)(a) = f^k(a) for k = 0,1,..., n-1 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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    (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.
    If you want something more accessible at A-level, I think integrating n times by parts is a better bet.
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    (Original post by DFranklin)
    If you want something more accessible at A-level, I think integrating n times by parts is a better bet.
    Ah yes, I'm not very familiar with it myself -- Analysis I basically steered clear of integration throughout till we actually got around to it.
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    Proof of Maclauren expansion using integration by parts.

    Assumption: f is n+1 times differentiable on [0,a].

    f(a) = f(0) + \int_0^a f'(x)\,dx, 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 x-a, not x (remember we can add an arbitrary constant when we integrate). So

    \eqalign {f(a) &= f(0) + [(x-a)f'(x)]_0^a - \int_0^a (x-a) f'(x)\\ &= f(0) + a f'(a) - \int_0^a (x-a)f''(x)\,dx}.

    From this point we just integrate by parts normally:

    f(a) = f(0) + a f'(a) - \left[\frac{(x-a)^2}{2} f''(a)\right]_0^a + \int_0^a \frac{(x-a)^2}{2} f^{(3)}(x)\,dx

    f(a) = f(0) + a f'(a) + \frac{a^2}{2} f''(a) + \left[\frac{(x-a)^3}{3!} f^{(3)}(a)\right]_0^a - \int_0^a \frac{(x-a)^3}{3!} f^{(4)}(x) \,dx

    etc until we get


    f(a) = f(0) + a f'(a) + \frac{a^2}{2} f''(a) + ... \frac{a^n}{n!}f^{(n)}(a) + \int_0^a \frac{(a-x)^n}{n!} f^{(n+1)}(x) \,dx

    (For a formal proof it's easier to start from this final expression and use induction, of course).
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    (Original post by DFranklin)
    Proof of Maclauren expansion using integration by parts.

    Assumption: f is n+1 times differentiable on [0,a].

    f(a) = f(0) + \int_0^a f'(x)\,dx, 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 x-a, not x (remember we can add an arbitrary constant when we integrate). So

    \eqalign {f(a) &= f(0) + [(x-a)f'(x)]_0^a - \int_0^a (x-a) f'(x)\\ &= f(0) + a f'(a) - \int_0^a (x-a)f''(x)\,dx}.

    From this point we just integrate by parts normally:

    f(a) = f(0) + a f'(a) - \left[\frac{(x-a)^2}{2} f''(a)\right]_0^a + \int_0^a \frac{(x-a)^2}{2} f^{(3)}(x)\,dx

    f(a) = f(0) + a f'(a) + \frac{a^2}{2} f''(a) + \left[\frac{(x-a)^3}{3!} f^{(3)}(a)\right]_0^a - \int_0^a \frac{(x-a)^3}{3!} f^{(4)}(x) \,dx

    etc until we get


    f(a) = f(0) + a f'(a) + \frac{a^2}{2} f''(a) + ... \frac{a^n}{n!}f^{(n)}(a) + \int_0^a \frac{(a-x)^n}{n!} f^{(n+1)}(x) \,dx

    (For a formal proof it's easier to start from this final expression and use induction, of course).
    Thank you for typing all of this out. In the third line, how do you get that the integral of (x-a) with respect to x is (x-a)^2 / 2?
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    (Original post by peterw55)
    Thank you for typing all of this out. In the third line, how do you get that the integral of (x-a) with respect to x is (x-a)^2 / 2?
    Standard result (for me). If you doubt it, differentiate (x-a)^2 /2.
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    (Original post by DFranklin)
    Standard result (for me). If you doubt it, differentiate (x-a)^2 /2.
    Yes, of course it works.

    x-a has antiderivative 0.5x^2 - ax (ignoring the +c)
    = x^2 / 2 - ax
    = x^2 / 2 - 2ax / 2
    = x^2 - 2ax / 2

    (x-a)^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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    (Original post by peterw55)
    Yes, of course it works.

    x-a has antiderivative 0.5x^2 - ax (ignoring the +c)
    = x^2 / 2 - ax
    = x^2 / 2 - 2ax / 2
    = x^2 - 2ax / 2

    (x-a)^2 / 1 = x^2 + a^2 - 2ax / 2
    What stupid algebra mistake have I made? Where does the a^2 / 2 come from?
    Arbitrary constant. Both (x^2 - 2ax) /2 and (x^2 - 2ax +a^2) /2 have derivative (x-a).
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    (Original post by DFranklin)
    Arbitrary constant. Both (x^2 - 2ax) /2 and (x^2 - 2ax +a^2) /2 have derivative (x-a).
    Because you can set a = 0?
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    (Original post by peterw55)
    Because you can set a = 0?
    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 anti-derivative.

    Similarly, the integral of x-a is not (x^2-2ax)/2, but (x^2-2ax/2) + C.

    You have effectively taken C = 0, I have taken it to equal a^2/2.
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    (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 anti-derivative.

    Similarly, the integral of x-a is not (x^2-2ax)/2, but (x^2-2ax/2) + C.

    You have effectively taken C = 0, I have taken it to equal a^2/2.
    Of course! Now I deservingly feel stupid :-/
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    (Original post by peterw55)
    Of course! Now I deservingly feel stupid :-/
    But I also now get your proof :-)

    Even if it's not intuitive, it's a good answer to my question, so thank you!
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    (Original post by peterw55)
    Even if it's not intuitive, it's a good answer to my question, so thank you!
    For an intuitive explanation, imagine we have two balls. If they start at the same place, with the same initial speed, and the same acceleration, and the same rate at which the acceleration changes, and the same rate at which the rate of change of the acceleration changes etc..., then it's intuitively believable that they will move in exactly the same way.

    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...

    Gotta love maths!
 
 
 
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