# About the remainder of Taylor expansion and Riemann-Liouville integral

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? (Start with equation 9.) –  Ｊ. Ｍ. May 8 '12 at 3:20
Maybe this interests you. –  Pedro Tamaroff May 8 '12 at 3:23
@J.M.,Thanks. Your reference is very helpful. But I have another question: Is $f(x)=I^{k+1}f^{(k+1)}$ true? –  Tim May 8 '12 at 3:53

Your first Question: $$f(x)=f(a)+\int_a^x{f'(t)}dt$$

$$\int{U'(t)V(t)}dt=U(t)V(t)-\int{U(t)V'(t)}dt$$

$U(t)=t-x$ ,$V(t)=f'(t)$

$$f(x)=f(a)+\int_a^x{f'(t)}dt=$$$$=f(a)+ [(t-x)f'(t)]_{t=a}^{t=x}-\int_a^x{(t-x)f''(t)}dt=$$$$= f(a)- (a-x)f'(a)-\int_a^x{(t-x)f''(t)}dt=$$$$=f(a)+ (x-a)f'(a)-[\frac{(t-x)^2}{2}f''(t)]_{t=a}^{t=x}+\int_a^x{\frac{(t-x)^2}{2}f'''(t)}dt=$$$$=f(a)+ (x-a)f'(a)-[\frac{(t-x)^2}{2}f''(t)]_{t=a}^{t=x}+\int_a^x{\frac{(t-x)^2}{2}f'''(t)}dt=$$$$=f(a)+ (x-a)f'(a)+\frac{(x-a)^2}{2}f''(a)+\int_a^x{\frac{(t-x)^2}{2}f'''(t)}dt$$

If you continue in that way you can get final result $$f(x)=\sum_{i=0}^k\frac{f^{(i)}(a)}{i!}(x-a)^i+\int_a^x\frac{f^{(k+1)}(t)}{k!}(x-t)^kdt$$

Riemann-Liouville integral is defined by $$I^{\alpha}[f(x)]=\frac{1}{\Gamma(\alpha)}\int_a^x{f(t)(x-t)^{(\alpha-1)}}dt$$ Equation (1)

Information about the integral from Wiki: where Γ is the Gamma function and a is an arbitrary but fixed base point. The integral is well-defined provided ƒ is a locally integrable function, and α is a complex number in the half-plane re(α) > 0. The dependence on the base-point a is often suppressed, and represents a freedom in constant of integration. Clearly I1ƒ is an antiderivative of ƒ (of first order), and for positive integer values of α Iαƒ is an antiderivative of order α by Cauchy formula for repeated integration

$$\Gamma (\alpha)=\int_{0}^{\infty }t^{\alpha-1}e^{-t}dt\tag{0}$$

if $\alpha$ positive integer

$$(\alpha-1)!=\Gamma (\alpha)=\int_{0}^{\infty }t^{\alpha-1}e^{-t}dt\tag{0}$$

if you replace $\alpha=k+1$ and $f(x)-->f^{(k+1)}(x)$ in Equation (1) $$I^{k+1}[f^{(k+1)}(x)]=\frac{1}{\Gamma(k+1)}\int_a^x{f^{(k+1)}(t)(x-t)^{k}}dt=\frac{1}{k!}\int_a^x{f^{(k+1)}(t)(x-t)^{k}}dt$$

$$f(x)=\sum_{i=0}^k\frac{f^{(i)}(a)}{i!}(x-a)^i+\int_a^x\frac{f^{(k+1)}(t)}{k!}(x-t)^kdt=\sum_{i=0}^k \frac{f^{(i)}(a)}{i!}(x-a)^i+I^{k+1}[f^{k+1}(x)]$$
$$I^{k+1}[f^{k+1}(x)]=f(x)-\sum_{i=0}^k \frac{f^{(i)}(a)}{i!}(x-a)^i$$
Thanks for your detailed explanation. But my problem is, in my imagination, Riemann-Liouville is high-order antiderivative. $I^1 f'(x)=f(x)$, $I^2 f''(x)=f(x)$. Intuitively, it seems natural to think $I^{k+1}f^{(k+1)}=f(x)$ rather than $I^{k+1}f^{(k+1)}=f(x)-$former k terms of Taylor series. –  Tim May 8 '12 at 11:55