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math equation and theory
original form of Taylor series
f(x)=f(a)+f′(a)1!(x−a)+f′′(a)2!(x−a)2+⋯
By making a substitution, we can find something that resembles the incremental form of the Taylor series:
let x−a=h, so x=h+a
f(x) = f(a+h)= f(a) + \frac{f'(a)}{1!}(h) + \frac{f''(a)}{2!}(h)^2+ \cdots \\
If we now replace a with x, we arrive at the incremental form of the Taylor series: f(x+h)=f(x)+f′(x)1!h+f′′(x)2!h2+⋯
It looks as if we're expanding at some unspecified point x instead of a and using h as an error term that will better approximate the value at f(x) as we use higher order terms. Could someone confirm whether this thinking is correct? This explanation makes sense to me intuitively, but I'm not sure.
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