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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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    @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

1 Answers 1

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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) $I^{k+1}[f^{(k+1)}(x)]=\frac{1}{\Gamma(k+1)}\int_a^x{f^{(k+1)}(t)(x-t)^{k}}dt=\frac{1}{k!}\int_a^x{f^{(k+1)}(t)(x-t)^{k}}dt$

Your second Question:

$f(x)=\sum_{i=0}^k\frac{f^{(i)}(a)}{i!}(x-a)^i+\int_a^x\frac{f^{(k+1)}(t)}{k!}(x-t)^kdt=\sum_{i=0}^k \frac{f^{(i)}(a)}{i!}(x-a)^i+I^{k+1}[f^{k+1}(x)]$

$I^{k+1}[f^{k+1}(x)]=f(x)-\sum_{i=0}^k \frac{f^{(i)}(a)}{i!}(x-a)^i$

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    quick question. if $U(t)=t-x$ ,$V(t)=f'(t)$ , then i would presume that f(t), rather than $f(x)=f(a)+\int_a^x{f'(t)}dt=$. let me know where i am wrong!2015-07-27