August 22, 2011

Taylor Series

The Taylor series for a (infinitely differentiable) function is a polynomial that approximates it. It can be represented in a few ways:
f(x)=f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+h.o.t., which is equivalent to \sum_{n=0}^\infty\frac{f^{(n)}(a)}{n!}(x-a)^n and can be rewritten as:
f(x+h)=f(x)+hf'(x)+\frac{h^2}{2!}f''(x)+h.o.t.

Each can be used to form Maclaurin series, where the derivatives are evaluated at zero. Some of the more common ones (such as e^x) are worth memorising so you can spot them when they come up, but most can be evaluated easily enough, as long as you remember how.

You can also expand for more than one variable by expanding once for each in turn until the desired level of accuracy is reached. A two-dimensional quadratic approximation is
f(x+h,y+k)=f(x,y)+h\frac{\partial f}{\partial x}(x,y)+k\frac{\partial f}{\partial y}(x,y)+\frac{1}{2}\left(h^2\frac{\partial^2f}{\partial x^2}(x,y)+2hk\frac{\partial^2f}{\partial x \partial y}(x,y)+k^2\frac{\partial^2f}{\partial y^2}(x,y)\right)+h.o.t.
which can be expanded to a general vector expansion as
f(\vec{x}+\vec{h})=f(\vec{x})+\vec{h}\cdot\nabla f(\vec{x})+\frac{1}{2}\sum_{1\leq i,j \leq n}h_ih_j\frac{\partial^2f}{\partial x_i \partial x_j}+h.o.t.
[Geometry and Motion notes]

Analysis II introduces Taylor’s Theorem, which explains just why Taylor Series work, and how good an approximation they are depending on how far you get into the sum.


- 2 comments by 0 or more people Not publicly viewable

  1. Sue

    Observe everything. Deduce. Eliminate the impossible, whatever remains, no matter how mad it might seem, must be the truth.

    23 Aug 2011, 08:41

  2. Nick

    Have you come across Cauchy’s terrible flat function?
    f(x)=exp(-1/x²) for non-zero x and f(0)=0
    At zero, it cons everyone into thinking it will never be positive (ALL derivatives are 0),
    and yet, somehow, it then goes up!
    I have read that Cauchy used this example to show that calculus cannot be developed
    algebraically, that using epsilons and deltas really is necessary (though I don’t quite
    see how that follows yet!)

    24 Aug 2011, 10:35


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