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If we get the Maclaurin expansion for $f(x)$, but we have an infinite number of terms, this would actually be the function $f(x)$(?) My thinking is that at $0$, the expansion would be the same as $f(x)$, and every other derivative would be equal so surely it would match $f(x)$ exactly?

Say $f(x)=\frac{1}{1-x}$, then the Maclaurin series is clearly a geometric progression, $1+x+x^2+x^3+...$, and for large $x$ we can see it obviously isn't remotely close to $f(x)$. So is this just a weird bit of maths, or is there something I'm missing, ie is my first paragraph wrong? Sorry if this is a bit confusing, I just can't get my head around why Maclaurin only works for certain $x$ values.

Botond
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mrmonkey99
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  • Because it's radius of convergence is $1$. – Botond Feb 26 '18 at 19:58
  • https://math.stackexchange.com/questions/1308992/why-doesnt-a-taylor-series-converge-always/1309167 The short version: Taylor polynomials have some guarantee of accuracy in a neighborhood of the point of expansion, but the neighborhood depends on the degree. If the derivatives are badly behaved, the neighborhood may potentially contract to just the point of expansion as the number of terms goes to infinity. In other cases, the neighborhood may grow (like for $\exp$); in still other cases, the neighborhood may stay about the same size (like for $1/(1-x)$). – Ian Feb 26 '18 at 20:00

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