calculus proof

Question

If f(x)=Σx⁴ⁿ/(4n)! ....prove that the 4th derivative equals f (x)

Answer 1

f(x)=Σx⁴ⁿ/(4n)!  One assumes the question is based on f(x)=Σ_n=0^∞x⁴ⁿ/(4n)!

f^iv(x)=Σ_n=0^∞ (4n)(4n-1)(4n-2)(4n-3)x^4n-4/(4n)!=Σ_n=4^∞ x^4n-4/(4n-4)! the terms with n<4 become zero. This sum is exactly f(x), as the same terms are summed in both cases.  In an expanded illustration we have

f(x)=1+x⁴/4!+x⁸/8!+x¹²/12!+...+x^4k/(4k)!+...

f'(x)=0+x³/3!+x⁷/7!+x¹¹/11!+...+x^4k-1/(4k-1)!+...

f"(x)=x²/2!+x⁶/6!+X¹⁰/10!+...+X^4K-2/(4K-2)!+...

...

f^iv(x)=1+x⁴/4!+x⁸/8!+...+x^4k-4/(4k-4)!+...=f(x)