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What is the Maclaurin series for? : #1/root(3)(8-x)#

Answer 1

Ezekiel Alexander

# 1/root(3)(8-x) = 1 + 1/48x + 1/576 x^2 + 7/41472x^3 + ... #
# |x| lt 8 #

We can derive a MacLaurin Series by using the Binomial Expansion: The binomial series tell us that:

# (1+x)^n = 1+nx + (n(n-1))/(2!)x^2 (n(n-1)(n-2))/(3!)x^3 + ...#

We can write:

# 1/root(3)(8-x) = 1/(8-x)^(1/3)# # " " = 1/((8)(1-x/8))^(1/3)# # " " = 1/2(1-x/8)^(-1/3)#

And so for the given function, we can replace "#x#" by #-x/8# and substitute #n=-1/3# in the binomial series:

# 1/root(3)(8-x) = 1/2{1 + (-1/3)(-x/8) + ((-1/3)(-4/3))/(2!)(-x/8)^2 + ((-1/3)(-4/3)(-7/3))/(3!)(-x/8)^3 + ... #

# " " = 1/2{1 + x/24 + (4/9)/(2) x^2/64 + (28/27)/(6)x^3/512 + ... } #

# " " = 1/2{1 + 1/24x + 1/288 x^2 + 7/20736x^3 + ... } #

# " " = 1 + 1/48x + 1/576 x^2 + 7/41472x^3 + ... #

The Radius of Convergence is given by:

# | -x/8 | lt 1 => |x| lt 8 #

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Answer 2

Cameron Alexander

The Maclaurin series is used to approximate functions as infinite polynomials centered at zero. The Maclaurin series expansion for ( \frac{1}{\sqrt{3}(8-x)} ) is ( \sum_{n=0}^{\infty} \frac{(8-x)^n}{\sqrt{3} \cdot 8^{n+1}} ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.