Set up expansion formula for Taylor Series #f(x)=\root(3)(x)# centered round #a=1#?

So far I have the exponent as #x^(-2/3-n)#, and the denominator as #3#. But the rest in the numerator?

What kind of factorial is #(1*2*5*8*....)#?

1 Answer
VNVDVI
May 10, 2018

#root(3)(x)=sum_(n=0)^oo(-1)^(n+1)(x-1)^n(2*5*8*11*...*(3n-1))/(n!3^n)#

Explanation:

#f(x)=x^(1/3)#

#f'(x)=1/3x^(-2/3)#

#f''(x)=(-1*2)/3^2x^(-5/3)#

#f'''(x)=(1*2*5)/3^3x^(-8/3)#

#f^((4))(x)=(-1*2*5*8)/3^4x^(-11/3)#

#f^((5))(x)=(1*2*5*8*11)/3^5x^(-14/3)#

We see that the derivatives alternate signs after the first term and the #n=1# term is positive, so, if we ignore the #n=0# term, we have a clear pattern represented by an instance of #(-1)^(n+1).#

Furthermore, the exponent can be represented by #x^(1/3-n)#.

The denominator involves #3^n.#

For the numerator, a distinct pattern emerges if we disregard #f(x)# and consider only its derivatives. That is, we strip out #f(1)=root(3)(1)=1# and start at #n=1.#

It can't be represented by a factorial (or even double factorial), but we can use the following notation to indicate the terms all have differences of #3# and alternate between even and odd:

We have #2*5*8*11*...*(3n-1)#

So,

#f^((n))(x)=(-1)^(n+1)x^(1/3-n)(2*5*8*11*...*(3n-1))/3^n#

#f^((n))(1)=(-1)^(n+1)1^(1/3-n)(2*5*8*11*...*(3n-1))/3^n#

#1^(1/3-n)=1#, so

#f^((n))(1)=(-1)^(n+1)(2*5*8*11*...*(3n-1))/3^n#

Thus, using the standard form of a Taylor Series about #a#,

#f(x)=sum_(n=0)^oof^((n))(a)(x-a)^n/(n!)# and stripping out the first term, we get

#root(3)(x)=1+sum_(n=1)^oo(-1)^(n+1)(x-1)^n(2*5*8*11*...*(3n-1))/(n!3^n)#

However, if we take the above summation at #n=0, # we get #-(-1)/(0!3^0)=1,# so, unintentionally, the #0th# term can be absorbed into the series!

#root(3)(x)=sum_(n=0)^oo(-1)^(n+1)(x-1)^n(2*5*8*11*...*(3n-1))/(n!3^n)#

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