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
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)