How to Calculate the Maclaurin Series for ( sqrt{1 x} )

The Maclaurin series is a powerful tool in calculus for approximating functions using polynomials. This article will walk through the process of calculating the Maclaurin series for the function ( f(x) sqrt{1 x} ).

Step-by-Step Derivation of the Maclaurin Series

To begin, let's use the Maclaurin series formula:

[ f(x) f(0) f'(0)x frac{f''(0)}{2!}x^2 frac{f'''(0)}{3!}x^3 cdots ]

Where ( f(x) sqrt{1 x} ). We need to find and evaluate the derivatives of ( f(x) ) at ( x 0 ).

Calculate ( f(0) ):

[ f(0)  sqrt{1   0}  1 ]

Calculate ( f'(x) ):

[ f'(x)  frac{1}{2sqrt{1   x}} ]

Evaluating at ( x 0 ):

[ f'(0)  frac{1}{2sqrt{1   0}}  frac{1}{2} ]

Calculate ( f''(x) ):

[ f''(x)  -frac{1}{4(1   x)^{3/2}} ]

Evaluating at ( x 0 ):

[ f''(0)  -frac{1}{4(1   0)^{3/2}}  -frac{1}{4} ]

Calculate ( f'''(x) ):

[ f'''(x)  frac{3}{8(1   x)^{5/2}} ]

Evaluating at ( x 0 ):

[ f'''(0)  frac{3}{8(1   0)^{5/2}}  frac{3}{8} ]

Calculate ( f^{(4)}(x) ):

[ f^{(4)}(x)  -frac{15}{16(1   x)^{7/2}} ]

Evaluating at ( x 0 ):

[ f^{(4)}(0)  -frac{15}{16(1   0)^{7/2}}  -frac{15}{16} ]

Summary of derivatives:

( f(0) 1 ) ( f'(0) frac{1}{2} ) ( f''(0) -frac{1}{4} ) ( f'''(0) frac{3}{8} ) ( f^{(4)}(0) -frac{15}{16} )

Substituting these values into the Maclaurin series formula, we get:

[ sqrt{1   x}  1   frac{1}{2}x - frac{1}{4}frac{x^2}{2!}   frac{3}{8}frac{x^3}{3!} - frac{15}{16}frac{x^4}{4!}   cdots ]

This simplifies to:

[ sqrt{1   x}  1   frac{1}{2}x - frac{1}{8}x^2   frac{1}{16}x^3 - frac{5}{64}x^4   cdots ]

Generalization for (sqrt{1 x^a})

If we generalize the function as ( f(x) (1 x)^{a} ) with ( a frac{1}{2} ), the derivatives become simpler to compute. Using the generalized binomial coefficient:

[ binom{a}{k} frac{a(a-1)cdots(a-k 1)}{k!} ]

We can write the Taylor series around ( x 0 ) as:

[ 1 x^a sum_{k0}^infty binom{a}{k} x^k ]

For the specific case ( a frac{1}{2} ), the binomial coefficient can be simplified as:

[ binom{frac{1}{2}}{k}  frac{frac{1}{2} left( frac{1}{2} - 1 right)cdots left( frac{1}{2} - k   1 right)}{k!}  frac{(-1)^{k-1}}{k!} frac{1}{2} frac{3}{2} cdots frac{2k-3}{2} ]

This simplifies further to:

[ binom{frac{1}{2}}{k}  frac{(-1)^{k-1}}{k!} frac{1}{2^k} frac{(2k-2)!}{2^{k-1} (k-1)!}  frac{(-1)^{k-1}}{k!} frac{1}{2^{2k}} frac{2k!}{2k-1} ]

The radius of convergence is determined by the ratio test, which shows the series converges for ( -1 x 1 ).

Conclusion

The Maclaurin series for ( sqrt{1 x} ) is a powerful representation that allows us to approximate the square root function using polynomials. This series is particularly useful in numerical analysis, physics, and engineering where polynomial approximations are needed. Understanding the process of finding Maclaurin series for more complex functions, such as generalizations involving ( (1 x)^a ), enhances our calculus toolkit.