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Integral form of Taylor expansion looks like this:$$f(x)=\sum_{i=0}^k\frac{f^{(i)}(a)}{i!}(x-a)^i+\int_a^x\frac{f^{(k+1)}(t)}{k!}(x-t)^kdt$$ Riemann-Liouville integral is $$I^{\alpha}f=\frac{1}{\Gamma(\alpha)}\int_a^x{f(t)(x-t)^{(\alpha-1)}}dt$$ Q1: The integral form remainder of Taylor expansion is exactly $I^{k+1}f^{(k+1)}$. Why is that?

Q2: As far as I know, Riemann-Liouville integral is basically $\alpha$th antiderivative of $f(x).$ Shouldn't $f(x)=I^{k+1}f^{(k+1)}$? Is that right?

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    Have you seen [this](http://mathworld.wolfram.com/TaylorSeries.html)? (Start with equation 9.)2012-05-08
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    Maybe [this](http://math.stackexchange.com/questions/112958/the-error-term-in-taylor-series-and-convolution) interests you.2012-05-08
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    @J.M.,Thanks. Your reference is very helpful. But I have another question: Is $f(x)=I^{k+1}f^{(k+1)}$ true?2012-05-08

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