Question

How do you find MacLaurin's Formula for `f(x)=sqrt(1+x)` and use it to approximate `f(1/2)` within 0.01?

Answer 1

`f(1.2) ~~ 1.21`

Explanation:

Let:

`f(x) = sqrt(1+x)`

The Maclaurin series is given by
`f(x) = f(0) + (f'(0))/(1!)x + (f''(0))/(2!)x^2 + (f'''(0))/(3!)x^3 + ... (f^((n))(0))/(n!)x^n + ...`

Although we could use this method, it is actually quicker, in this case, to use a Binomial Series 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 + ...`

And so for the given function, we have:

`f(x) = (1+x)^(1/2)`

`\ \ \ \ \ \ \ = 1 + (1/2)x + (1/2(-1/2))/(2!)x^2 + (1/2(-1/2)(-3/2))/(3!)x^3 + (1/2(-1/2)(-3/2)(-5/2))/(4!)x^4 + ...`

`\ \ \ \ \ \ \ = 1 + 1/2x - (1/4)/(2)x^2 + (3/8)/(6)x^3 - (15/16)/(24)x^4 + ...`

`\ \ \ \ \ \ \ = 1 + 1/2x - 1/8x^2 + 1/16x^3 - 5/128x^4 + ...`

And hopefully we have sufficient terms to answer the remainder of the question with sufficient accuracy!

To estimate `f(1/2)`, we put `x=1/2`, and truncate terms, giving:

`f(1/2) ~~ 1 + 1/2(1/2) - 1/8(1/2)^2 + 1/16(1/2)^3 - 5/128(1/2)^4 + ...`
`\ \ \ \ \ \ \ \ \ \ \ = 1 + 1/4 - 1/8(1/4) + 1/16(1/8) - 5/128(1/16) + ...`

`\ \ \ \ \ \ \ \ \ \ \ = 1 + 1/4 - 1/32 + 1/128 - 5/2048 + ...`

`\ \ \ \ \ \ \ \ \ \ \ = 9983/8224`

`\ \ \ \ \ \ \ \ \ \ \ = 1.2132886`

`\ \ \ \ \ \ \ \ \ \ \ = 1.21` to 2dp, as requested

Noting that as `5/2048 =0.00244\ « \ 0.01`

For comparison with the actual answer. we have:

`f(1/2) = sqrt(1+1/2) = sqrt(3/2) = 1.224744 ...`