Question

[3] Prove that: Let \( m-1<\alpha<m \) and \( f \) be continuously diff. Function , then there exists \( \xi \in(a, b) \) such that \[ \frac{f(b)-\sum_{k=0}^{m-1} \frac{(b-a)^{k}}{k!} f^{(k)}(a)}{(b-a)^{\alpha}}=\frac{{ }^{c} D_{x}^{\alpha} f(\xi)}{\Gamma(\alpha+1)} \]

          [3] Prove that:
Let \( m-1<\alpha<m \) and \( f \) be continuously diff. Function , then there exists \( \xi \in(a, b) \) such that
\[
\frac{f(b)-\sum_{k=0}^{m-1} \frac{(b-a)^{k}}{k!} f^{(k)}(a)}{(b-a)^{\alpha}}=\frac{{ }^{c} D_{x}^{\alpha} f(\xi)}{\Gamma(\alpha+1)}
\]

[3] Prove that:
Let m-1<α<m and f be continuously diff. Function , then there exists ξ∈(a, b) such that

(f(b)-∑k=0^m-1((b-a)^k)/(k!) f^(k)(a))/((b-a)^α)=(^c Dx^α f(ξ))/(Γ(α+1))

Added by Evelyn G.

Calculus

Dale Varberg, Edwin Purcell deceased,… 9th Edition

Chapter 3

Instant Answer

Solved by Expert Scott Stetson

Step 1

The equation involves a fractional derivative, which is a generalization of the usual derivative to non-integer orders. Show more…

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[3] Prove that: Let \( m-1<\alpha<m \) and \( f \) be continuously diff. Function , then there exists \( \xi \in(a, b) \) such that \[ \frac{f(b)-\sum_{k=0}^{m-1} \frac{(b-a)^{k}}{k!} f^{(k)}(a)}{(b-a)^{\alpha}}=\frac{{ }^{c} D_{x}^{\alpha} f(\xi)}{\Gamma(\alpha+1)} \]